Linear Algebra problems im having

joey12345

hi everyone im having a few problems with some linear algebra. my exams coming up this month and im feeling strong enough just hopefuly some of you guys can clear a few things up for me. heres the problems thanks in advance.
Q1
if a=
2 -1
-1 2
b=
1 0
0 3

show that the most general 2x2 matrix M such that AM=MB is of the form
x y
x -y.

Q2.
if A=
0 1 0
0 0 1
1 2 -1

find constants p,q and r such that A^3=pA^2+qA+rI.

thats read, A cubed = p by A squared + q by A + r by I(indentity matrix)

Q3.

If

At is the transpose of the square matrix A, then A is symmetric if At = A and is skew-symmetric if At = -A. Show that A+ At is symmetric and A- At is skew-symmetric.

Express the matrix

A=
2 1 3
-2 0 1
3 1 2

as the sum of a symmetric matrix and a skew-symmetric matrix.



more problems might arise as i get more invloved in the study. by the way this is not homework or anything and im not being lazy these are examples taken from previous year exam papers so any solutions and explanations would be great.

thanks,
Joe

LeixlipRed

Well you won't be getting any solutions. Explanations on how you might solve them yourself perhaps.

So, tell us how you think you might solve the first one for example?

joey12345

hey thanks for the reply. if people could explain to me how its done thats even better than the solutions so i can understand it better. in regard to how id go about it im not too sure im sure its very simple if i knew what the most general matrix M means?
as you can prob tell linear algebra is definitely not my strong point but how and ever.

LeixlipRed

Most general is a slightly odd way to put it. If you had to call M something? In general what would you call it if you didn't know any of the entries?

LeixlipRed

Once you've given M entries in general you then want to show that AM and MB have the same entries in the left column and the right column entries differ by a minus.

joey12345

i would probably say that
m=
a b
c d

or something to that effect with 4 unknowns?

LeixlipRed

That's the job, now sub it into that equation and multiply out both sides, what do you have then? 2 matrices that equal each other so what do we know about their entries?

joey12345

ok so i subbed in
m=
a b
c d

on both sides.

when i multiplied out i got
2a-c 2b-d
-a+2c -b+2d
=
a 3b
c 3d

LeixlipRed

But what do you know about the corresponding entries then? If both sides are equal?

joey12345

hmmm i cant really see anything else from where i last was? its prob something very easy im just not seeing it.

LeixlipRed

If two things are equal they must be identical. So if two matrices are equal then they must be identical. So the first entry in the first row must equal the first entry in the first row on the other side of the equality, see what I mean?

joey12345

yea i get ya but i thought you were saying that they are equal and i was gettin confused. so can i go any further with the problem or is the answer that they are not equal?

LeixlipRed

Have a read of the original problem again. You want to find M in it's "most general" form, essentially it's simplest form. So if you know 2a-c=a what do you know about the relationship between a and c?

Saying they are not equal is not true, they are equal for when certain equations like that one are satisfied. And those equations come from setting the corresponding entries equal to each other.

If I asked you about an equation x=2y, you would not say the left is not equal to the right. You would tell me that x must be twice y. See what I mean?

joey12345

kk i got it. when i simplify each corresponding equation like 2a-c=a we get c=a and then when you do -a+2c=c you get a=c obviously. and then with the other two you get -d=b and -b=d which can obviously be wrote as d=-b and the other stays the same so therefore M in its most general form CAN be wrote as
x y
x -y

correct?

LeixlipRed

Excellent! Good job, what about the next one then?

joey12345

thanks a mil i no that was very simple its just i havent done linear algebra in over a year because my first year in college i failed linear algebra cause i never went to the class (stupid move) and then ive had to take a year out because my college wanted me to pay an outstanding amount of dosh just to sit the module for the year again so ive just been at home studying so i can into second year in september. anyways yea ill go for the second one. any pointers?

Michael Collins

joey12345 wrote: »

hi everyone im having a few problems with some linear algebra. my exams coming up this month and im feeling strong enough just hopefuly some of you guys can clear a few things up for me. heres the problems thanks in advance.
Q1
if a=
2 -1
-1 2
b=
1 0
0 3

show that the most general 2x2 matrix M such that AM=MB is of the form
x y
x -y.

You should have enough help from Leixlip with this one.

Q2.
if A=
0 1 0
0 0 1
1 2 -1

find constants p,q and r such that A^3=pA^2+qA+rI.

thats read, A cubed = p by A squared + q by A + r by I(indentity matrix)

Check out the Caley-Hamilton theorem, and apply it here.

Q3.

If

At is the transpose of the square matrix A, then A is symmetric if At = A and is skew-symmetric if At = -A. Show that A+ At is symmetric and A- At is skew-symmetric.

Express the matrix

A=
2 1 3
-2 0 1
3 1 2

as the sum of a symmetric matrix and a skew-symmetric matrix.



more problems might arise as i get more invloved in the study. by the way this is not homework or anything and im not being lazy these are examples taken from previous year exam papers so any solutions and explanations would be great.

thanks,
Joe

For the proof, use the fact that

[latex] \displaystyle (\mathbf{A}+\mathbf{B}) ^\mathrm{T} = \mathbf{A}^\mathrm{T} + \mathbf{B}^\mathrm{T} [/latex]

Then apply this fact to the matrix A.

If you've any questions just ask.

Michael Collins

Michael Collins wrote: »

Check out the Caley-Hamilton theorem, and apply it here.

Or you could just sub in A and solve it like you've now done with the last one I suppose. The first way is more elegant but you may not have heard of the Cayley-Hamilton theorem.

LeixlipRed

I'm sure the intention of the lecture was to use the theorem so you should try applying it at least

joey12345

hi guys. these are questions from previous exam papers so as long as it doesnt state that ive to use the theory i dont think i have to but thanks anyways. i subbed in A and dis is what ive gotten so far.
0 1 0
0 0 1
1 8-1
=
0 P 0
0 0 P
P 4P P

+

0 Q 0
0 0 Q
Q 2Q -Q

+

R 0 0
0 R 0
0 0 R


I have a feeling i did something wrong here?

LeixlipRed

Looks fine to me. Now what can you do with the 3 matrices on the right hand side of the equality? They're all of equal dimensions so you can add them together.

joey12345

k i got p=1, q=2, r=0?
what i did was add up the entries so i had
r , p+q , 0
0 , r , p+q
p+q , 4p+2q , p-q+r

(commas are just there so u no which element corresponds to which column)

so i had 4p+2q=8 which became 2p+q=4

i then had p-q+r=-1

i knew r=0 from the matrix.

so i had a silmultaneous eq.

4p+2q=8
p-q=-1

and den solved for p and q

correct?

Michael Collins

joey12345 wrote: »

hi guys. these are questions from previous exam papers so as long as it doesnt state that ive to use the theory i dont think i have to but thanks anyways. i subbed in A and dis is what ive gotten so far.
0 1 0
0 0 1
1 8-1
=
0 P 0
0 0 P
P 4P P

+

0 Q 0
0 0 Q
Q 2Q -Q

+

R 0 0
0 R 0
0 0 R


I have a feeling i did something wrong here?

Unfortunately raising matrices to powers isn't quite that simple! Probably just a slip on your part but [latex] \mathbf{A^3} = \mathbf{A} \cdot \mathbf{A} \cdot \mathbf{A} [/latex], which is not the same as raising the individual elements of the matrix to the power of 3.

Hint: [latex] A^3 = \left[ \matrix{1 & 2 & -1\cr -1 & -1 & 3\cr 3 & 5 & -4} \right]
[/latex]

That hint is for doing it your way. But the Cayley-Hamilton way is actually a lot faster. Find the Characteristic Polynomial, then post what ya get and I'll show you how easy it is from there!

Hint 2: [latex] \hbox{Characteristic Polynomial} = \hbox{det}(\mathbf{A}-I\lambda) [/latex]

LeixlipRed

Oops, didn't check your work!

joey12345

oooops dunno what i was thinking when raising the matrice. ill do it all out with subbing A in. ill get the characteristic polynomial aswell and i can see the two ways being done.

joey12345

hey k when i subbed in i ended up with \left [ \matrix{1 & 2 & -1\cr -1 & -1 & 3\cr 3 & 5 &-4} \right]= \left [ \matrix{r & q & p\cr p & 2p+r & -p+q\cr -p+q & -p+2q & 3p-q+r} \right]

so i saw r=1 q=2 and p= -1

so just to be sure i subbed in the values for
3p-q+r=-4

and they satisfied the equation and as far as the characteristic polynomal i used the equation

\hbox{Characteristic Polynomial} = \hbox{det}(\mathbf{A}-I\lambda)

i ended up with

-lambda^3-lambda^2+2lambda+1

i dunno if i did the characteristic polynomial correct but let me no when ya get a chance

joey12345

i tried to write the matrix and the formula like you were doing so if they dont show up correctly let us no and ill edit the post so its readable thanks

Michael Collins

joey12345 wrote: »

hey k when i subbed in i ended up with [latex] \left [ \matrix{1 & 2 & -1\cr -1 & -1 & 3\cr 3 & 5 &-4} \right]= \left [ \matrix{r & q & p\cr p & 2p+r & -p+q\cr -p+q & -p+2q & 3p-q+r} \right] [/latex]

so i saw r=1 q=2 and p= -1

so just to be sure i subbed in the values for
3p-q+r=-4

and they satisfied the equation and as far as the characteristic polynomal i used the equation

[latex] \hbox{Characteristic Polynomial} = \hbox{det}(\mathbf{A}-I\lambda) [/latex]

i ended up with

-lambda^3-lambda^2+2lambda+1

i dunno if i did the characteristic polynomial correct but let me no when ya get a chance

You just need to put the "[ latex ]" tag before your maths and finish it with "[ /latex ]" - but leave out the spaces between the brackets and "latex", and leave out the quotes. (For some reason stuff I put in "[ CODE ]" tags gets parsed by the server...). I've done this when I quoted your post above anyway.

So the Characteristic Polynomial you got (which is correct) was:

[latex] \displaystyle -\lambda^3 -\lambda^2 +2\lambda + 1 [/latex]

but the Cayley-Hamilton Theorem says that every square matrix satisfies i.e. solves, its own Characteristic Polynomial. So your matrix [latex] {\bf A} [/latex] must solve this equation:

[latex] \displaystyle -{\bf A}^3 - {\bf A}^2 +2{\bf A}+{\bf I} = 0[/latex]

so

[latex] \displaystyle {\bf A}^3 = -{\bf A}^2 +2{\bf A}+{\bf I}[/latex]

so comparing this to the given expression to get:

[latex] \displaystyle -{\bf A}^2 +2{\bf A}+{\bf I} = p{\bf A}^2 +q{\bf A}+r{\bf I}[/latex]

we can see that

[latex]
\eqalign{
p &= -1\cr
q &= 2\cr
r &= 1}
[/latex]

which is exactly what you got by subbing in. So you can be confident you're correct.

joey12345

jesus that is much easier. ive been looking through the exam papers there and there isnt anymore that are similar to that one there when else can the cayley hamilton theory be applied ? can it really only be applied when theres unknowns like that one?

joey12345

hi micheal collins

have another problem here to do with vectors. its on the exam paper and wondering if youve any more tips and tricks on how to solve them

show that the line joining P(2, -1, 4) and Q(-3, 5, 0) does not intersect the plane 2x+y-z=8

LeixlipRed

Probably a million ways of doing that one! If two objects don't intersect each other what would you expect to find if you attempted to find their points of intersection?

Or a vector on the line, what would it's relationship to the normal vector of the plane be if they don't intersect? Think about the relationship the line and the plane must have if they don't intersect....

joey12345

LeixlipRed wrote: »

Probably a million ways of doing that one! If two objects don't intersect each other what would you expect to find if you attempted to find their points of intersection?

Or a vector on the line, what would it's relationship to the normal vector of the plane be if they don't intersect? Think about the relationship the line and the plane must have if they don't intersect....

hi leixlipred thanks for the reply. if the line and the plane dont intersect then they must have different coordinates. and if two objects dont intersect then they must not have any coordinates that are the same?

p.s thanks for the help with the "most general" matrix question. they show up multiple times each year so thanks to you ive hopefully got a little headstart with the exam next week

LeixlipRed

Yeh, that's kind of the general idea but the second way I mentioned is easier to compute. Do you know what the normal vector to a plane is? Try draw a picture, a plane and a line, what is the only type of line that won't touch a plane? One that is parallel to the plane! Otherwise it will pass through it. So if the plane is parallel to the line the normal vector of that plane and a vector on that line will be....?

joey12345

LeixlipRed wrote: »

Yeh, that's kind of the general idea but the second way I mentioned is easier to compute. Do you know what the normal vector to a plane is? Try draw a picture, a plane and a line, what is the only type of line that won't touch a plane? One that is parallel to the plane! Otherwise it will pass through it. So if the plane is parallel to the line the normal vector of that plane and a vector on that line will be....?

when a line and a plane are parallel then they are the same distance from eachother always. a normal vector is perpendicular to any other vector on the plane but i dont know the relationship from the normal vector to the vector on the parallel line? is it that theyll always be perpendicular ?

LeixlipRed

Yeh, if a line and plane are parallel then the normal vector of the plane will be perpendicular to any vector on or parallel to that line. Now, how do you show two vectors are perpendicular?

joey12345

LeixlipRed wrote: »

Yeh, if a line and plane are parallel then the normal vector of the plane will be perpendicular to any vector on or parallel to that line. Now, how do you show two vectors are perpendicular?

two vectors are perpendicular if there dot product is equal to zero.

LeixlipRed

Exactly so do you know what you have to do now?

joey12345

LeixlipRed wrote: »

Exactly so do you know what you have to do now?

just looking on yotube there for a few examples do i get the vector equation of the line?

or is it much more simpler than this and im just not seeing it?

LeixlipRed

The vector equation of a line won't be a vector, it'll be an equation. All you need is to get a vector on that line. You have two points on the line, how do you do that?

joey12345

LeixlipRed wrote: »

The vector equation of a line won't be a vector, it'll be an equation. All you need is to get a vector on that line. You have two points on the line, how do you do that?

i just take away Q from P so ill end up with <1, 6, -4>

and then i get the dot product with the normal vector of the plane.

i take it that if the the plane is this..... 2x+y-z=8

then the normal vector is <2, 1, -1>

i dont think thats right is it?

because when i get there dot products its a non zero answer and that would mean that there not perpindicular?

LeixlipRed

You've calculated QP wrong, check your work.

joey12345

LeixlipRed wrote: »

You've calculated QP wrong, check your work.

ahhhh man sorry didnt see that Q had a MINUS 3.

yea that worked out fine and dandy thanks for the help leixlipred.

say if that question was if they DO intersect?

would it be done a different way?

what you think of this one?

show that the vectors
a= i + 2j - 3k
b=2i - j + 2k
c=3i + j -k

lie in the same plane?(ie coplanar)

so is all i have t do here is find the determinant? and if it equals zero then it they are co planar?

joey12345

k got that one i just got the determinant which was equal to 0 which means the vectors are linearly dependent and therefore coplanar.

Michael Collins

joey12345 wrote: »

ahhhh man sorry didnt see that Q had a MINUS 3.

yea that worked out fine and dandy thanks for the help leixlipred.

say if that question was if they DO intersect?

would it be done a different way?

what you think of this one?

show that the vectors
a= i + 2j - 3k
b=2i - j + 2k
c=3i + j -k

lie in the same plane?(ie coplanar)

so is all i have t do here is find the determinant? and if it equals zero then it they are co planar?

Just thinking for the original problem you should probably also check that the line doesn't lie in the plane, as if it did technically it would intersect it everywhere, but still be perpendicular to the normal to the plane. It's easy enough to do this for a single point.

joey12345

i was actually thinking that myself right ill check that now so.

joey12345

right i plugged in the values and none of them satisfied the equation. thats right isnt it? just plug in one of the points into the equation?

or do i use the parametric representation?

LeixlipRed

Michael Collins wrote: »

Just thinking for the original problem you should probably also check that the line doesn't lie in the plane, as if it did technically it would intersect it everywhere, but still be perpendicular to the normal to the plane. It's easy enough to do this for a single point.

Ah good spot, forgot to mention that. Was trying to think what is the quickest way of approaching this question. I reckon if you find the parametric equation of the line and plug it into the plane to get a contradiction is another quick way of doing it. No matter though, all methods are valid.

joey12345

LeixlipRed wrote: »

Ah good spot, forgot to mention that. Was trying to think what is the quickest way of approaching this question. I reckon if you find the parametric equation of the line and plug it into the plane to get a contradiction is another quick way of doing it. No matter though, all methods are valid.

yea i done the parametric representation there and when i was solving for T it cancelled itself out which would mean that it doesnt intersect the plane right?

is another way to do it just subbing in P or Q?

like for example P(2, -1, 4). could i use 2=x, -1=y, and 4=z and then sub in into the plane to get a contradiction? or is that wrong? well either way the dot product worked out and so did the parametric representation.

LeixlipRed

Just subbing in the point proves nothing other than that point doesn't lie on the plane. A line is a collection of infinitely many points so no that method does not answer the original question.

As for subbing in the parametric equations into the plane, the reason this shows that our line and plane intersect is because we cannot find a t that satisfies this equation as you alluded to.

joey12345

Michael Collins wrote: »

Just thinking for the original problem you should probably also check that the line doesn't lie in the plane, as if it did technically it would intersect it everywhere, but still be perpendicular to the normal to the plane. It's easy enough to do this for a single point.

LeixlipRed wrote: »

Just subbing in the point proves nothing other than that point doesn't lie on the plane. A line is a collection of infinitely many points so no that method does not answer the original question.

As for subbing in the parametric equations into the plane, the reason this shows that our line and plane intersect is because we cannot find a t that satisfies this equation as you alluded to.

sorry i wasnt clear there thats what i mean by subbing in P that will prove that its not on the plane if you get a contradiction.

LeixlipRed

Yeh you prove the point P is not on the plane. But what about the line joining P to Q? It tells you nothing about that.