How to get the two eigen vectors for eigen =1

Question

I have to find the eigen vectors for this matrix. \begin{pmatrix} 1 & 0 & 1\\ 0 & 1 & 0\\ 0 & 0 & 0 \end{pmatrix}

I end up with this matrix to plug in the eigen values. \begin{pmatrix} 1-\lambda & 0 & 1\\ 0 & 1-\lambda & 0\\ 0 & 0 & -\lambda \end{pmatrix}

Eigen values are 1,0

When I plug in zero then I get \begin{pmatrix} -1\\ 0\\ 1 \end{pmatrix}

But I'm not sure how the answer key is getting two vectors when using eigen = 1 The vectors should be: \begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix}

\begin{pmatrix} 0\\ 1\\ 0 \end{pmatrix}

Answer 1

For each eigenvalue $\lambda$, use the definition of "eigenvector" and just solve

$\begin{pmatrix} 1 & 0 & 1\\ 0 & 1 & 0\\ 0 & 0 & 0 \end{pmatrix}\begin{pmatrix} a \\ b \\c \end{pmatrix}=\lambda \begin{pmatrix} a \\ b \\c \end{pmatrix}$.

This will furnish you with eigenvectors for eigenvalues. It turns out in this case that the eigenspace for the value $1$ is two dimensional, so you can produce two linearly independent eigenvectors for $1$.

From $\begin{pmatrix} 1 & 0 & 1\\ 0 & 1 & 0\\ 0 & 0 & 0 \end{pmatrix}\begin{pmatrix} a \\ b \\c \end{pmatrix}=\begin{pmatrix} a \\ b \\c \end{pmatrix}$

we learn that $\begin{pmatrix} a +c \\ b \\0 \end{pmatrix}=\begin{pmatrix} a \\ b \\c \end{pmatrix}$, so in other words $c=0$, and an eigenvector for $1$ must look like this:

$\begin{pmatrix} a \\ b \\0 \end{pmatrix}$. Now you just have to find two different sets of coefficients satisfying this such that the vectors are linearly independent: pretty easy to do in this case. You already have an answer in front of you, so see how it fits in.

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This is equivalent to substituting the eigenvalue into the lambda matrix, and then computing generators of the nullspace of that matrix.

In this case, for $\lambda=1$, you would be looking for solutions to this equation:

$\begin{pmatrix} 0 & 0 & 1\\ 0 & 0 & 0\\ 0 & 0 & -1 \end{pmatrix}\begin{pmatrix} a \\ b \\c \end{pmatrix}= \begin{pmatrix} 0 \\ 0 \\0 \end{pmatrix}$.

The resulting thing you learn is that $ \begin{pmatrix} c \\ 0 \\-c \end{pmatrix}= \begin{pmatrix} 0 \\ 0 \\0 \end{pmatrix}$, which is just saying $c=0$.

That means you are free to choose $a,b$ in the vector $ \begin{pmatrix} a \\ b \\0 \end{pmatrix}$ as you wish, just as in my previous paragraphs.

Answer 2

If $\lambda$ is an eigenvalue for a matrix $A$, then an eigenvector for $\lambda$ is a vector $v$ such that $Av = \lambda v$, or $(A-\lambda I)v = 0$. So the eigenvectors corresponding to $\lambda$ are exactly those vectors whose image under $A-\lambda I$ is the zero vector --- that is, they are just the nullspace of $A-\lambda I$. Presumably you already know how to calculate the nullspace of a matrix.

Answer 3

If we have a matrix $A=(a_{ij})_{1\le i,j\le n}$ in the standard basis of $\Bbb R^n$ so we have

$$A e_j=\sum_{i=1}^n a_{ij}e_i$$ In our case we can read immediately from the given matrix that $$Ae_1=e_1\quad;\quad Ae_2=e_2\quad\text{and}\quad Ae_3=e_1$$ so we see that $e_1$ and $e_2$ are eigenvectors associated to the eigenvalue $1$. Of course the last eigenvalue is $0$ and we see easily that $e_1-e_3$ is an eigenvector associated to it and $A$ is similar to the diagonal matrix $\operatorname{diag}(1,1,0)$.