0
$\begingroup$

$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?

  • 0
    I get \left[\matrix{-3&2&0\cr 0&-2&3\sqrt2\cr 0&0&0} \right] as a reduced form.2011-12-05

1 Answers 1

1

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.)

  • 1
    ${2\over 3}R_1=[ -2\ \ {4\over 3}\ \ 0] $; ${2\over3}R_1+R_2 =[ 0\ \ -{2\over3}\ \ \sqrt2]$.2011-12-05