Eigenspaces, Orthogonal + Diagonal Matrices

**kabirSarin** · May 14th 2010, 10:24 AM

Could someone please let me know how I would go about accomplishing the following tasks. Thank you.

Let A = [[4,0,0],[0,1,3],[0,3,1]

a) Find the Eigenvalues of A.

(lambda - 4)(lambda - 4)(lambda + 2)
So the eigenvalues are 4, 4, and -2.

b) For each eigenvalue, find a basis for the corresponding eigenspace.

??

c) Use these bases vectors to construct an orthogonal matrix P.

??

d) Verify that $P^-1AP$ is a diagonal matrix.

??

**dwsmith** · May 14th 2010, 12:32 PM

Originally Posted by kabirSarin

Could someone please let me know how I would go about accomplishing the following tasks. Thank you.

Let A = [[4,0,0],[0,1,3],[0,3,1]

a) Find the Eigenvalues of A.

(lambda - 4)(lambda - 4)(lambda + 2)
So the eigenvalues are 4, 4, and -2.

b) For each eigenvalue, find a basis for the corresponding eigenspace.

??

c) Use these bases vectors to construct an orthogonal matrix P.

??

d) Verify that $P^-1AP$ is a diagonal matrix.

??

Eigenvalues of A are

$\begin{bmatrix} 4-\lambda & 0 & 0\\ 0 & 1-\lambda & 3\\ 0 & 3 & 1-\lambda \end{bmatrix}\rightarrow (4-\lambda)[(1-\lambda)^2-9]=-(\lambda-4)^2(\lambda+2)$

When $\lambda=4$

$\begin{bmatrix} 0 & 0 & 0\\ 0 & -3 & 3\\ 0 & 3 & -3 \end{bmatrix}\rightarrow rref= \begin{bmatrix} 0 & 0 & 0\\ 0 & 1 & -1\\ 0 & 0 & 0 \end{bmatrix}$

$x_1$
$x_2=x_3$
$x_3$

$\begin{bmatrix} x_1\\ x_3\\ x_3 \end{bmatrix}\rightarrow x_1\begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}+x_3\begin{bmatrix} 0\\ 1\\ 1 \end{bmatrix}$

Now find the other eigenvector for lambda = -2

Once you have the third eigenvector, you can construct your P matrix that corresponds to A

**HallsofIvy** · May 16th 2010, 03:42 AM

I prefer to use the definition of "eigenvector" directly: If 4 is an eigenvector for A, then there exist a vector, v, such that Av= 4v. Specifically,
$\begin{bmatrix}4 & 0 & 0 \\ 0 & 1 & 3 \\ 0 & 3 & 1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}4x \\ y+ 3z\\ 3y+ z\end{bmatrix}= 4\begin{bmatrix} x \\ y \\ z\end{bmatrix}$ .

That gives the four equations 4x= 4x, y+ 3z= 4y, and 3y+ z= 4z. The first equation is satisfied by any x, of course and the last two are the same as 3y= 3z or z= y. That is, any eigenvector, corresponding to eigenvalue 4, is of the form $\begin{bmatrix}x \\ y \\ y\end{bmatrix}= x\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}+ y\begin{bmatrix}0 \\ 1 \\ 1\end{bmatrix}$ , showing the two basis vectors for the eigenspace clearly.

With eigenvalue -2, those equations are simply 4x= -2x, y+ 3z= -2y, and 3y+ z= -2z. The first equation says x= 0 and I will leave the others to you.

Thread: Eigenspaces, Orthogonal + Diagonal Matrices

LinkBack

Thread Tools

Display

Eigenspaces, Orthogonal + Diagonal Matrices

Similar Math Help Forum Discussions

Proving that the product of orthogonal matrices is also orthogonal.

General solution for diagonal zero matrices?

Field of diagonal matrices (or not...)

Linear maps and diagonal matrices

Diagonal Matrices from Characteristic Poly

Search Tags