Eigenspaces, Orthogonal + Diagonal Matrices

May 2010
1
0
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 \(\displaystyle P^-1AP\) is a diagonal matrix.

??
 

dwsmith

MHF Hall of Honor
Mar 2010
3,093
582
Florida
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 \(\displaystyle P^-1AP\) is a diagonal matrix.

??
Eigenvalues of A are

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

When \(\displaystyle \lambda=4\)

\(\displaystyle \begin{bmatrix}
0 & 0 & 0\\
0 & -3 & 3\\
0 & 3 & -3
\end{bmatrix}\rightarrow rref= \begin{bmatrix}
0 & 0 & 0\\
0 & 1 & -1\\
0 & 0 & 0
\end{bmatrix}\)

\(\displaystyle x_1\)
\(\displaystyle x_2=x_3\)
\(\displaystyle x_3\)

\(\displaystyle \begin{bmatrix}
x_1\\
x_3\\
x_3
\end{bmatrix}\rightarrow x_1\begin{bmatrix}
1\\
0\\
0
\end{bmatrix}+x_3\begin{bmatrix}
0\\
1\\
1
\end{bmatrix}\)

Now find the other eigenvector for lambda = -2

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

MHF Helper
Apr 2005
20,249
7,909
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,
\(\displaystyle \begin{bmatrix}4 & 0 & 0 \\ 0 & 1 & 3 \\ 0 & 3 & 1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}4x \\ y+ 3z\\ 3y+ z\end{bmatrix}= 4\begin{bmatrix} x \\ y \\ z\end{bmatrix}\).

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 \(\displaystyle \begin{bmatrix}x \\ y \\ y\end{bmatrix}= x\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}+ y\begin{bmatrix}0 \\ 1 \\ 1\end{bmatrix}\), 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.