# [SOLVED]Eigenvectors and Eigenvalues (311.5.5.15)

#### karush

##### Well-known member
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*}

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

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Staff member
That is correct.

#### karush

##### Well-known member
\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)

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

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#### Opalg

##### MHB Oldtimer
Staff member
\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.

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#### karush

##### Well-known member
that was a great help

I have more to post