Question

If the eigenvalues of matrix T are 2,2,3 how do you find 3 linearly independent eigenvectors

Answer 1

What is matrix \(T\)?

Answer 2

T= 5 3 -3
1 3 -1
3 3 -1

Answer 3

Thanks. Same post with subbed values:
\[T=\begin{pmatrix}5&3&-3\\1&3&-1\\3&3&-1\end{pmatrix}\]
The eigenvalues were obtained by finding \(\lambda\) such that
\[|T-\lambda I|=0,\]
or
\[\begin{vmatrix}5-\lambda&3&-3\\1&3-\lambda&-1\\3&3&-1-\lambda\end{vmatrix}=0\]
You're given the eigenvalues, \(\lambda_1=\lambda_2=2\) and \(\lambda_3=3\).

To find the corresponding eigenvectors, you must find the vectors \(\vec{\eta}_1,\vec{\eta}_2,\vec{\eta}_3\) that satisfy the following matrix equations:
\[\begin{cases}
T\vec{\eta}_1=\vec{0}\\
T\vec{\eta}_2=\vec{\eta}_1&\text{because 2 is a repeated eigenvalue}\\
T\vec{\eta}_3=\vec{0}
\end{cases}\]

Answer 4

So now you're left with solving for those vectors.
\[\begin{pmatrix}3&3&-3\\1&1&-1\\3&1&-1\end{pmatrix}\begin{pmatrix}\eta_1\\\eta_2\\\eta_3\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}\]
Row reduce the leftmost matrix:
\[\begin{pmatrix}3&3&-3&|&0\\1&1&-1&|&0\\3&1&-1&|&0\end{pmatrix}~~\Rightarrow~~\begin{pmatrix}1&0&0&|&0\\0&1&-1&|&0\\0&0&0&|&0\end{pmatrix}\]
This tells you that \(\eta_1=0\) and \(\eta_2=\eta_3\). Choose any non-zero values for \(\eta_2\) and \(\eta_3\). I'd pick 1 for convenience.

Thus, \(\vec{\eta}_1=\begin{pmatrix}0\\1\\1\end{pmatrix}\).

Answer 5

Let me know if anything doesn't make sense so far.

Answer 6

I see your row reduction but how did you get vector 0 1 1

Answer 7

Writing the row-reduced augmented matrix as a system of equations might make it clearer to see:
\[\begin{cases}\eta_1+0\eta_2+0\eta_3=0\\
0\eta_1+\eta_2-\eta_3=0\\
0\eta_1+0\eta_2+0\eta_3=0\end{cases}\]
From the first equation, you get \(\eta_1=0\). That's the only way the first equation works.

The second equation tells you that \(\eta_2=\eta_3\). Now, any value for these \(\eta\)'s will work, so long as this equation is satisfied. Using 1 makes things simple.

Answer 8

oh okay that makes sense and then how do you find the other eigenvectors

Answer 9

Oops, made a typo!
\[\begin{pmatrix}3&3&-3\\1&1&-1\\3&3&\color{red}{-3}\end{pmatrix}\begin{pmatrix}\eta_1\\\eta_2\\\eta_3\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}\]
Row reduce:
\[\begin{pmatrix}3&3&-3&|&0\\1&1&-1&|&0\\3&3&-3&|&0\end{pmatrix}~~\Rightarrow~~\begin{pmatrix}1&1&-1&|&0\\0&0&0&|&0\\0&0&0&|&0\end{pmatrix}\]
This tells you that \(\eta_1+\eta_2-\eta_3=0\). This equation solutions that will provide linearly independent eigenvectors, so you can ignore the \(T\vec{\eta}_2=\vec{\eta}_1\) note I made.

Some solutions that work might be \(\vec{\eta}_1=\begin{pmatrix}1\\0\\1\end{pmatrix}\) and \(\vec{\eta}_2=\begin{pmatrix}1\\-1\\0\end{pmatrix}\).

Answer 10

so those two vectors are for lambda=2

Answer 11

Correct. The third vector will correspond to \(\lambda_3=3\).
\[\begin{pmatrix}2&3&-3\\1&0&-1\\3&3&-4\end{pmatrix}\begin{pmatrix}\eta_1\\\eta_2\\\eta_3\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}\]
Row reduce:
\[\begin{pmatrix}2&3&-3&|&0\\1&0&-1&|&0\\3&3&-4&|&0\end{pmatrix}~~\Rightarrow~~\begin{pmatrix}1&0&-1&|&0\\0&1&-\frac{1}{3}&|&0\\0&0&0&|&0\end{pmatrix}\]
First row tells you that \(\eta_1=\eta_3\). The second row says that \(3\eta_2=\eta_3\).

For a convenient vector, I'd choose \(\eta_2=1\). This makes \(\eta_1=\eta_3=3\), so \(\vec{\eta}_3=\begin{pmatrix}3\\1\\3\end{pmatrix}\).

Answer 12

yep that's what I got thanks!

Answer 13

yw