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I'm looking to diagonalize a matrix A seen below. (Find a $P$ and a $D$ such that $AP = PD$). $$ \begin{bmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 1 & 2 & 3 & 4 & 5 & 6 \\ 1 & 2 & 3 & 4 & 5 & 6 \\ 1 & 2 & 3 & 4 & 5 & 6 \\ 1 & 2 & 3 & 4 & 5 & 6 \\ 1 & 2 & 3 & 4 & 5 & 6 \\ \end{bmatrix} $$

It doesn't seem practical to use determinants in this matrix to find the eigenvalues. I do realize that with row reduction I can limit to the matrix:

$$ \begin{bmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ \end{bmatrix} $$

And further realize that the following vector is in the column space of A.

$$ \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ \end{bmatrix} $$

By solving $Av_1=λv_1$ I get an eigenvalue of 21. But this is the only eigenvalue that I get and I believe it has a multiplicity of 1. Then is there a $λ=0$ with multiplicity 5? Further, how is it possible to find the matrix $P$ (specifically the eigenvectors that form a basis for the eigenspace associated with $λ=0$)?

econom
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2 Answers2

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We need to find the eigenvectors $\mathbf{x}$ such that: $A\mathbf{x} =\mathbf{0}.$ Because the system is homogeneous, we can use the row echelon form of $A$, thus we need to solve the sytem: $$ \begin{bmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ \end{bmatrix} \cdot \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\x_ 6 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0\end{bmatrix},$$ Thus, we have the equation: $$x_1 + 2x_2 + 3x_3 + 4x_4 + 5x_5 + 6x_6 = 0,$$ which implies that we have $5$ free variables.

So, every eigenvector corresponding to the zero eigenvalue can be written in the form: $$x_2\cdot\begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \\0 \\0 \end{bmatrix} +x_3 \cdot \begin{bmatrix} -3 \\ 0 \\ 1 \\ 0 \\0 \\0 \end{bmatrix} +x_4\cdot \begin{bmatrix} -4 \\ 0 \\ 0 \\ 1 \\0 \\0 \end{bmatrix}+x_5\cdot\begin{bmatrix} -5 \\ 0 \\ 0 \\ 0 \\1 \\0 \end{bmatrix} +x_6\cdot \begin{bmatrix} -6 \\ 0 \\ 0 \\ 0 \\0 \\1 \end{bmatrix}.$$

The column vectors form an eigenbasis for the eigenspace that corresponds to the zero eigenvalue. So, we have the matrix $$P = \begin{bmatrix} 1 & -2 & -3 & -4 & -5 & -6\\ 1 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}$$ and $D = \operatorname{diag}(21\quad 0\quad 0\quad 0\quad 0\quad 0 ).$

We can notice that matrix $A$ is a rank one matrix, so there is only one non-zero eigenvalue, which can be found by adding the diagonal elements of $A$.

thanasissdr
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  • Thank you - this does make a lot of sense. Where I seem to get most hung up here is calculating the first eigenvalue. Typically I'd find the determinant of (A-λI) and the associated characteristic polynomial which doesn't seem very doable here. This seems like a special case and I'm not sure how applicable it is for other diagonalization problems? I'm also unsure why we'd assume λ = 21 only has a multiplicity of 1 and λ = 0 has multiplicity 5? – econom May 16 '16 at 00:18
  • Yes - in a typical diagonalization problem I would find the characteristic polynomial (from determinant(A-λI)). That doesn't seem practical here because det(A-λI) would be a huge hassle to calculate by hand. I can get $λ_1 = 21$ from recognizing that $Av_1=λv_1$, but how do we know it only has multiplicity 1? And that $λ=0$ has multiplicity of 5? – econom May 16 '16 at 00:39
  • Very nice! Thanks for the trick. Again, appreciated with your help, @thanasissdr! – econom May 16 '16 at 01:20
  • @econom The matrix has rank $1$, so $0$ is an eigenvalue with multiplicity $5$, leaving the non-zero eigenvalue to have multiplicity one. Another was to see this is that the image is one-dimensional, so there’s at most a single line that’s mapped to itself by $A$, i.e., that the eigenspace corresponding to the non-zero eigenvalue is one-dimensional. – amd May 16 '16 at 03:15
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You can immediately see that your matrix decomposes as a product $CR$ of a column matrix $C=(1~1~1~1~1~1)^\top$ and a row matrix $R=(1~2~3~4~5~6)$, so it has rank$~1$. For rank $1$ matrices, the analysis is easy: it has an eigenvalue $0$ with geometric multiplicity$~n-1$ (here $5$) and one more eigenvalue equal to the trace $t=RC$ of the matrix (here $t=21$); see this question and links from there for details. In case that $t=0$ the algebraic multiplicity of the eigenvalue$~0$ is strictly larger than its geometric multiplicity, and the matrix therefore non-diagonalisable; but if (as it is here) the trace is nonzero, then the matrix is diagonalisable.

The eigenspace for $0$ is the kernel of the matrix$~R$ (since vectors killed by $R$ are also killed by $CR$), which here is the subspace $\{(a~b~c~d~e~f)^\top\,\mid a+2b+3c+4d+5e+6f=0\,\}$. The eigenspace for the remaining eigenvalue$~t$ (if nonzero) is spanned by the column of $C$, since $CRC=Ct=tC$.

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Marc van Leeuwen
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