-Eigenvectors and Eigenvalues (311.5.5.15)

[karush]

Find a basis for eigenspace corresponding to the listed eigenvalue:
just seeing if these first steps are correct

\(\displaystyle \begin{align*}
A_{15}&=\left[
\begin{array}{rrr} -4&1&1\\ 2&-3&2\\ 3&3&-2 \end{array}
\right],\lambda=-5&(1)\\
A-(-5)i&=\left[
\begin{array}{rrr} -4&1&1\\ 2&-3&2\\ 3&3&-2 \end{array}
\right]-
\left[
\begin{array}{rrr} -5&0&0\\ 0&-5&0\\ 0&0&-5
\end{array}\right]=&(2)\\
&=\left[
\begin{array}{rrr} 1&1&1\\ 2&2&2\\ 3&3&3 \end{array}
\right]&(3)
\end{align*}\)$$\tiny{311.05.01.15;
Linear Algebra \, and \, its \, Applications; \, David \, C Lay; \, 4th \,Edition}$$

 

[Opalg]

That is correct.

 

[karush]

\(\displaystyle \begin{align*}
A_{15}&=\left[
\begin{array}{rrr} -4&1&1\\ 2&-3&2\\ 3&3&-2 \end{array}
\right],\lambda=-5&(1)\\
A-(-5)i&=\left[
\begin{array}{rrr} -4&1&1\\ 2&-3&2\\ 3&3&-2 \end{array}
\right]-
\left[
\begin{array}{rrr} -5&0&0\\ 0&-5&0\\ 0&0&-5
\end{array}\right]=&(2)\\
&=\left[
\begin{array}{rrr} 1&1&1\\ 2&2&2\\ 3&3&3 \end{array}
\right]&(3)\\ \\ \\
&=\left[\begin{array}{r}
-1\\ 1\\ 0
\end{array} \right]
\large\cdot
\left[\begin{array}{r}
-1\\ 1\\ 0
\end{array} \right]&(9)
\end{align*}\)

ok (9) is the answer but don't I know the steps between (3) and (9)

$$\tiny{311.05.01.15;
Linear Algebra \, and \, its \, Applications; \, David \, C Lay; \, 4th \,Edition}$$

 

[Opalg]

\(\displaystyle \begin{align*}
A-(-5)i&=\left[
\begin{array}{rrr} 1&1&1\\ 2&2&2\\ 3&3&3 \end{array}
\right]&(3)\\ \\...\\ \\
&=\left[\begin{array}{r}
-1\\ 1\\ 0
\end{array} \right]
\large\cdot
\left[\begin{array}{r}
-1\\ 1\\ 0
\end{array} \right]&(9)
\end{align*}\)

ok (9) is the answer but don't I know the steps between (3) and (9)

The eigenspace consists of the vectors \(\displaystyle \begin{bmatrix}x\\y\\z\end{bmatrix}\) such that \(\displaystyle (A+5I)\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}0\\0\\0\end{bmatrix}.\) So you need to solve the system of equations $$\begin{bmatrix}1&1&1 \\ 2&2&2 \\ 3&3&3 \end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}0\\0\\0\end{bmatrix}.$$ That system reduces to the single equation $x+y+z = 0.$ To find a basis for the subspace consisting of the vectors satisfying that equation, one way would be to say that $y$ and $z$ can be arbitrary but then $x$ must satisfy $x = -y-z.$ Then $$\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}-y-z\\y\\z\end{bmatrix} = y\begin{bmatrix}-1\\1\\0\end{bmatrix} + z\begin{bmatrix}-1\\0\\1\end{bmatrix}.$$ So one possible basis for that subspace consists of the vectors $\begin{bmatrix}-1\\1\\0\end{bmatrix}$ and $\begin{bmatrix}-1\\0\\1\end{bmatrix}$, and that is what your equation (9) ought to say.

 

[karush]

that was a great helpI have more to post