What is the basis of $\lambda=4$ for this matrix?

$$A-4I= \left(\begin{array}{rrr} -3&2&0 \\ 2&-2&\sqrt2 \\ 0&\sqrt2&-3 \end{array}\right)$$

From my calculations it seems that $x_1,x_2 \text{ and }x_3$ are all leading variables. However, my teacher expressed the basis in terms of $x_3$. What I know is that we can't express a basis in terms of leading variables, we can only in terms of free variables. Any help please?

The reduced form is: $\left(\begin{array}{rrr} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}\right)$

So all three are linearly independent, right?

How am I supposed to construct the eigenspace of this?

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I'm not sure what you want to happen here. In your first equation, there is a $\lambda$ on the left, but not on the right... – J. M. Dec 5 '11 at 16:53
@J.M.it's not on the right, because $\lambda$ was 4 and I already subtracted it. – Andrew Dec 5 '11 at 16:55
Um, you can't. You must have made an arithmetic error somewhere. (And, on the left, do you want $A-4I$?) – David Mitra Dec 5 '11 at 16:55
@DavidMitra Yes, it's -4I. I corrected it. I'm trying to re-reduce it – Andrew Dec 5 '11 at 16:57
@andrew Because, if there weren't, then you wouldn't have obtained $I$ as the echelon form. – David Mitra Dec 5 '11 at 17:01

1 Answer

If you know that $\lambda=4$ is, indeed, an eigenvalue, then you would know the null space of the matrix $A-4I$ is non-trivial. There is a non-zero vector $\bf x$ with $(A-4I){\bf x}=\bf 0$.

Now the claimed reduced form of $A-4I$ clearly has a trivial null space. What a quandry...

We can say:

Either $\lambda=4$ is not an eigenvalue

or you made an arithmetic mistake in deriving the reduced form.

You need to check what's wrong. Is $4$ really an eigenvalue? If so, then you need to redo your reduction.

(I did it and obtained $\left[\matrix{-3&2&0\cr 0&-2&3\sqrt2\cr 0&0&0} \right]$, which has non-trivial solutions.)

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$\left(\begin{array}{rrr} -3&2&0 \\ 2&-2&\sqrt2 \\ 0&\sqrt2&-3 \end{array}\right)$ $\frac{2}{3}R_1+R_2$->$R_2$ $\left(\begin{array}{rrr} -3&2&0 \\ 0&-4&\sqrt2 \\ 0&\sqrt2&-3 \end{array}\right)$ $\frac{\sqrt2}{4}R_2+R_3$->$R_3$ $\left(\begin{array}{rrr} -3&2&0 \\ 0&-4&\sqrt2 \\ 0&0&\frac{-5}{2} \end{array}\right)$ This is what I get, I can't find the error – Andrew Dec 5 '11 at 17:18
${2\over 3}R_1=[ -2\ \ {4\over 3}\ \ 0]$; ${2\over3}R_1+R_2 =[ 0\ \ -{2\over3}\ \ \sqrt2]$. – David Mitra Dec 5 '11 at 17:22