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I want to diagonalize the following matrix:

$$A =\begin{bmatrix} 1&0&0&0 \\ 0&0&1&0 \\ 0&1&0&0\\ 0&0&0&1 \end{bmatrix}$$

And I get the following characteristic equation: $(\lambda - 1)^3(\lambda +1)$ and therefore $\lambda=1,-1$

Firstly get the basis vectors for $\lambda = 1$ $$(A - \lambda_2 I_4) = \begin{bmatrix} 0&0&0&0 \\ 0&-1&1&0 \\ 0&1&-1&0\\ 0&0&0&0 \end{bmatrix} \rightarrow \text{row reduce} \rightarrow \begin{bmatrix} 0&1&-1&0 \\ 0&0&0&0 \\ 0&0&0&0\\ 0&0&0&0 \end{bmatrix}$$

So I have $x_2 = x_3$ wich yields the vector: $$v_1 = \begin{bmatrix} 0 \\ 1 \\ 1\\ 0 \end{bmatrix}$$

However there should be $3$ basis vectors total for $\lambda = 1$ according to the characteristic equation.

The answer key gives vectors: $$\begin{bmatrix} 1 \\ 0 \\ 0\\ 0 \end{bmatrix} , \begin{bmatrix} 0 \\ 0 \\ 0\\ 1 \end{bmatrix}$$ to be the other basis vectors for $\lambda = 1$, but where do these vectors come from in regards to my reduced matrix? I understand there ought to be 2 additional vectors to the one that I found but why are they the vectors given?

theGuy05
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3 Answers3

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When you find eigenvectors of a matrix $A$ that corresponds to an eigenvalue $\lambda$

$$Av=\lambda v$$

or equivalently

$$\left(A-\lambda I\right)v=0$$

you need to find the basis of the null space of $A-\lambda I$, i.e.

$${\rm null}\left(A-\lambda I\right)=\left\{v|\left(A-\lambda I\right)v=0\right\}$$

You can easily see that in your case $e_{1}=\begin{pmatrix}1\\0\\0\\0\end{pmatrix}$ and $e_{2}=\begin{pmatrix}0\\0\\0\\1\end{pmatrix}$ indeed satisfy this criterion for the matrix $A-I$

$$\left(A-I\right)e_{1,2}=0$$

so these are also eigenvectors with $\lambda=1$.

eranreches
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You’ve found a constraint on $x_2$ and $x_3$, but there are no constraints whatsoever on $x_1$ and $x_4$, so all vectors in this eigenspace are of the form $(a,b,b,c)^T$. Can you see where the other two basis vectors came from now?

amd
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  • You might also want to have a look at this for a detailed description of how to read the basis for the null space from a rref matrix. – amd Dec 09 '17 at 22:16
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I guess you're in math251 since i just finished the same question! If you examine your matrix, you see that $x_1, x_3$ and $x_4$ are 'free' variables. The number of vectors you're going to end up with are therefore = to the number of free variables. so for $x_1$ you'll have $(1, 0, 0, 0)$, you already found $x_3$ and $x_4$ is $(0, 0, 0, 1)$

Badam Baplan
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Simon
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  • Nope, sounds like a different/class school. I'm guessing this is just one of those practice problems that are used all over. – theGuy05 Dec 09 '17 at 22:33