# Thread: Finding Eigenvectors to diagonalize

1. ## Finding Eigenvectors to diagonalize

Hello, I've been looking at question which states to diagonalize the matrix $A = \left(\begin{array}{ccc}1&t&0\\0&t&t\\0&0&0\end{ar ray}\right)$
I've found the eigenvalues to be 1, 0 and t. I know that if t = 1, the matrix is not diagonalizable, and if t = 0, then A is already diagonalized, and thus is diagonalizable. Now, I have to look at what happens when t is not equal to 1 or 0.
Looking at each eigenvalue, I'm having trouble finding the eigenvectors.
For $\lambda = 1$, they say, "Obviously $Ae_{1}=e_{1}$, therefore $e_{1}$ is a eigenvector." I don't follow that. It's not obvious to me.
For $\lambda = 0$, I know that this is the same as just finding Ax = 0, however, my eigenvector is $\left(\begin{array}{c}1/t\\-1\\1\end{array}\right)$, & the one the chose was $\left(\begin{array}{c}t\\-1\\1\end{array}\right)$. I can't see where I went wrong.
For $\lambda = t$, they have the eigenvector being $\left(\begin{array}{c}t\\t-1\\0\end{array}\right)$ I'm completely lost there.

How are all these eigenvectors found?
Thanks for your time.

2. Originally Posted by Silverflow
Hello, I've been looking at question which states to diagonalize the matrix $A = \left(\begin{array}{ccc}1&t&0\\0&t&t\\0&0&0\end{ar ray}\right)$
I've found the eigenvalues to be 1, 0 and t. I know that if t = 1, the matrix is not diagonalizable, and if t = 0, then A is already diagonalized, and thus is diagonalizable. Now, I have to look at what happens when t is not equal to 1 or 0.
Looking at each eigenvalue, I'm having trouble finding the eigenvectors.
For $\lambda = 1$, they say, "Obviously $Ae_{1}=e_{1}$, therefore $e_{1}$ is a eigenvector." I don't follow that. It's not obvious to me.
Then practice more! "Obviously", $Ae_1= \left(\begin{array}{ccc}1&t&0\\0&t&t\\0&0&0\end{ar ray}\right)\begin{pmatrix}1 \\ 0 \\ 0\end{pmatrix}= \begin{pmatrix}1 \\ 0 \\ 0 \end{pmatrix}$.

The whole point of being an eigenvalue is that there exist a non-zero vector, v, such that $Av= \lambda v$.

In this example, that would be $Av= \left(\begin{array}{ccc}1&t&0\\0&t&t\\0&0&0\end{ar ray}\right)\begin{pmatrix}x \\ y \\ z\end{pmatrix}= \lambda\begin{pmatrix}x \\ y \\ z\end{pmatrix}$ or, after multiplying on the left side of the equation, $\begin{pmatrix}x+ ty \\ ty+ tz \\ 0\end{pmatrix}= \begin{pmatrix} \lambda x \\ \lambda y \\ \lambda z\end{pmatrix}$

And so there exist x, y, z, not all 0, satisfying $x+ ty= \lambda x$, $ty+ tz= \lambda y$, and $0= \lambda z$.

If $\lambda= 1$, those equations become x+ ty= x, ty+ tz= y, and 0= z. From the last, z= 0 so the second equation becomes ty= y or (t-1)y= 0. Since $t\ne 1$, y= 0. The first equation then is x= x which is satisfied for all x. Any eigenvector corresponding to eigenvalue 1 must be of the form $\begin{pmatrix}x \\ 0 \\ 0\end{pmatrix}= x\begin{pmatrix} 1 \\ 0 \\ 0\end{pmatrix}$ which is spanned by $\begin{pmatrix}1 \\ 0 \\ 0\end{pmatrix}= e_1$.

For $\lambda = 0$, I know that this is the same as just finding Ax = 0, however, my eigenvector is $\left(\begin{array}{c}1/t\\-1\\1\end{array}\right)$, & the one the chose was $\left(\begin{array}{c}t\\-1\\1\end{array}\right)$. I can't see where I went wrong.
Why do you think you "went wrong"? With $\lambda= 0$, the equations, $x+ ty= \lambda x$, $ty+ tz= \lambda y$, and $0= \lambda z$, become x+ ty= 0, ty+ tz= 0, and 0= 0. From the first, ty= -x so the second is -x+ tz= 0 or x= tz. Then y= -x/t and z= x/t. An eigenvector corresponding to eigenvalue 0 is $\begin{pmatrix}x \\-x/t \\ x/t \end{pmatrix}= x\begin{pmatrix}1 \\ -1/t\ \\ 1/t \end{pmatrix}$. If you take x= 1, this is $\begin{pmatrix}1 \\ -1/t \\ 1/t\end{pmatrix}$. If you take x= t, this is $\begin{pmatrix} t\\ -1 \\ 1\end{pmatrix}$. Each is a multiple of the other and both are perfectly good eigenvectors. Any multiple of an eigenvector is an eigenvector. More generally, the set of all eigenvectors corresponding to the same eigenvalue is a subspace.

For $\lambda = t$, they have the eigenvector being $\left(\begin{array}{c}t\\t-1\\0\end{array}\right)$ I'm completely lost there.
With $\lambda= t$, the equations, $x+ ty= \lambda x$, $ty+ tz= \lambda y$, and $0= \lambda z$, become $x+ ty= tx$, $ty+ tz= ty$, and $0= tz$. Both the second and third equations, since $t\ne 0$, give z= 0. The first equation is (t-1)x= ty or $y= \frac{t-1}{t}x$. Any eigenvector corresponding to eigenvalue t is of the form $\begin{pmatrix}x \\ \frac{t-1}{t}x \\ 0\end{pmatrix}= x\begin{pmatrix}1 \\ \frac{t-1}{t}\\ 0\end{pmatrix}$. Taking x= t gives $\begin{pmatrix} t \\ t-1 \\ 0\end{pmatrix}$.

How are all these eigenvectors found?
Thanks for your time.

3. Thank you, for all that. I think these questions are a bit beyond me. So I shall revised the basics again before attempting then again. Thanks!