I'm confused on how to approach this problem. Given matrix A and the Identity I where A+I is row equivalent to

[1,-2,3,5][0,0,1,4][0,0,0,0][0,0,0,0]

Find the basis for the eigenspace E=-1. Thanks in advance for your help.

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- Sep 24th 2012, 10:45 AMnivek0078solving for basis of a given eigenspace
I'm confused on how to approach this problem. Given matrix A and the Identity I where A+I is row equivalent to

[1,-2,3,5][0,0,1,4][0,0,0,0][0,0,0,0]

Find the basis for the eigenspace E=-1. Thanks in advance for your help. - Sep 24th 2012, 11:30 AMSironRe: solving for basis of a given eigenspace
You probably know that the eigenvalues of map with matrix representation $\displaystyle A$ can be found by solving the equation $\displaystyle \det(A - \lambda I) = 0$ for $\displaystyle \lambda$. It appaered that $\displaystyle \lambda = -1$ is an eigenvalue therefore to find the corresponding eigenspace you'll have to find the nullspace of the matrix $\displaystyle A+I$. Can you do that first?

- Sep 24th 2012, 05:15 PMnivek0078Re: solving for basis of a given eigenspace
Well that's where I'm unsure of the problem gives the row reduced form of the matrix A+I. I don't know the original form of matrix A or how to solve for the vector.

- Sep 24th 2012, 05:31 PMTheEmptySetRe: solving for basis of a given eigenspace
Siron has given you everything that you need. As they said to find the eigen values of a matrix you solve the euqation.

$\displaystyle \det(A-\lambda I)=0$

This will give you the eigen values of A.

To find the eigen vectors you sub the eigen values back into the matrix equation $\displaystyle A-\lambda I$.

In your case you are told that $\displaystyle \lambda =-1$. Plug this into the matrix equation to get

$\displaystyle A-\lambda I = A-(-1)I=A+I$.

Can you finish from here? - Sep 24th 2012, 05:40 PMHallsofIvyRe: solving for basis of a given eigenspace
Ax= -x is the same as Ax+ x= (A+I)x= 0. So you need to find all x, y, z, u, that satisfy

$\displaystyle \begin{bmatrix}1 & -2 & 3 & 5 \\ 0 & 0 & 1 & 4 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}\begin{bmatrix}x \\ y \\ z \\ u\end{bmatrix}= \begin{bmatrix}0 & 0 & 0 & 0 \end{bmatrix}$ - Sep 25th 2012, 06:04 AMnivek0078Re: solving for basis of a given eigenspace
I want to thank you all for input! I'm sorry that not really seeing the answer right off. I've been at this kind of work for way too long. Anyways this is what i've got, I took the given matrix shown in the posting above.

-found the free and basic variables: x1,x3 are basic and x2,x4 are free

-did the system of equations and got this as the answer

[x1,x2,x3,x4]=[2x2-3x3-5x4,-x1,-3x3-5x4,x3,x4] = x3[-3,-3,1,0] + x4[-5,-5,0,1]

If this is not correct please tell me where i'm going wrong. Do you have to solve for the given matrix by plugging in the -1 for the lamda value allong the diagonals and then do the system of equations?