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The following matrix $A$ has $\lambda=2$ and $\lambda=8$ as its eigenvalues

$$ A = \begin{bmatrix} 4 & 2 & 2 \\ 2 & 4 & 2 \\ 2 & 2 & 4 \end{bmatrix}$$

let $P$ be the orthogonal matrix that diagonalizes $A$. in other words, $A= PDP^T$. you can check that

$$ P = \begin{bmatrix} (-√2)/2 & (-√6)/6 & (-√3)/3 \\ (√2)/2 & (-√6)/6 & (√3)/3 \\ 0 & (√6)/3 & (√3)/3 \end{bmatrix} $$

then the linear space of eigenvectors associated to the eigenvalue $\lambda=2$ is generated by?

  1. The last two columns of the matrix $P$
  2. The last two rows of matrix $P$
  3. The first two columns of the matrix $P$
  4. The first two rows of the matrix $P$
  5. The last column of the matrix $P$

I Don't know how to solve this, I tried doing it but it got really messy and I didn't get any answer that made sense. If someone could please explain how to do it or how to solve it, that would be great.

Thank You

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  • $\begingroup$ To format a matrix from now on use \begin{bmatrix} and \end{bmatrix}. In between separate the entries in the same row by & and separate the rows by \\. $\endgroup$
    – Benjamin
    Dec 13, 2014 at 20:45

2 Answers 2

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Since matrix $P$ is orthogonal ($P^\top = P^{-1}$) one can rewrite $A = PDP^\top$ as $$ D = P^{-1}A(P^{-1})^\top = P^\top AP $$ After multiplying one would obtain (there is a typo in your $P$ matrix) $$ D = \begin{bmatrix} -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ -\frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{6}} & \sqrt{\frac{2}{3}} \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \end{bmatrix} \begin{bmatrix} 4 & 2 & 2 \\ 2 & 4 & 2 \\ 2 & 2 & 4 \end{bmatrix} \begin{bmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\ 0 & \sqrt{\frac{2}{3}} & \frac{1}{\sqrt{3}} \end{bmatrix} = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 8 \end{bmatrix} $$ Note that $A = PDP^\top$ could be rewritten as $AP = PD$ which means that $P$ columns are eigenvectors of $A$ and corresponding eigenvalues are on $D$ diagonal: $$ A\mathbf{p}_i = d_{ii} \mathbf{p}_i, \qquad P = \begin{bmatrix} \mathbf{p}_1 & \mathbf{p}_2 & \mathbf{p}_3 \end{bmatrix} $$ Because $d_{11} = d_{22} = 2$ the eigenspace corresponding to $\lambda = 2$ is formed by $\mathbf{p}_1$ and $\mathbf{p}_2$

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The term is "eigenspace for the eigenvalues $\lambda=2$" (not a linear space of eigenvectors, since the zero vector is not an eigenvector). It is by definition the kernel of (the linear map with matrix) $A-2I$, which is a matrix with all entries $2$. A column vector is in this kernel if each row of $A-2I$ applied to it is zero, and since all three rows are equal, it suffices to consider just one. If $x,y,z$ are the entries of the column vector, the equation becomes $x+y+z=0$. The first two columns of $P$ satisfy this, the third does not.

In fact $P$ has apparently been formed by taking bases of the eigenspaces for $\lambda=2$ ($2$-dimensional) and for $\lambda=8$ ($1$-dimensional), and making each of these basis orthonormal. The rows of $P$ have no natural interpretation as vectors expressed on the standard basis (and indeed none of them satisfy the relation $x+y+z=0$).

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