# Find orthonormal basis and describe the reduced matrix

Let $$M$$ be an $$(n\times n)$$-matrix with eigenvector $$v=(1,2,\ldots,n)$$. Suppose all other eigenvectors of $$M$$ are orthogonal to $$v$$ and we are interested in finding these eigenvectors (and its corresponding eigenvalues).

Therefore, I want to "delete" the direction spanned by $$v$$ and reduce to the orthogonal part. This should give a matrix of one dimension less which is maybe easier to analyze to find the remaining eigenvectors. But how can one do this reduction?

Is it possible to determine an (in best case, orthonormal) basis of the orthogonal complement of the span of $$v$$ (this is of dimension $$n-1$$) and to reduce the matrix $$M$$ to this orthogonal complement? Can the new matrix, which is one dimension less, be described in terms of the orthonormal basis?

I really have no idea.

For example, let $$n=4$$ and $$M=\begin{pmatrix}-31 & -1 & -1 & 24\\ 58 & -62 & -2 & 48\\ -3 & 87 & -93 & 72\\ -4 & -4 & 116 & -24\end{pmatrix}$$

This matrix has eigenvalues $$\lambda_i$$ with corresponding eigenvectors $$v_i$$ (here they are known but in my situation the aim is to find the eigenvectors orthogonal to $$(1,2,3,4)$$; but for illustration let them be given):

• $$\lambda_1=60, v_1=(1,2,3,4)$$
• $$\lambda_2=-120, v_2=(-2,-4,-6,7)$$
• $$\lambda_3=-60, v_3=(-7,-8,-3,8)$$
• $$\lambda_4=-90, v_4=(-19,-38,-27,44)$$

The eigenvalues $$v_2,v_3,v_4$$ are orthogonal to $$v_1=(1,2,3,4)$$.

How can I find a (orthonormal) basis for $$\textrm{span}(v_1)^\perp$$, the orthogonal complement of the span of $$v_1$$ (which is 3-dimensional), and how can I get the matrix which is the restriction of $$M$$ to this 3d orthogonal complement and has the eigenvectors $$v_{2},v_3,v_4$$?

I think one basis for $$\textrm{span}(v_1)^\perp$$ is the set $$\left\{\begin{pmatrix}4\\0\\0\\-1\end{pmatrix},\begin{pmatrix}1\\0\\1\\-1\end{pmatrix},\begin{pmatrix}0\\2\\0\\-1\end{pmatrix}\right\}$$

For example, can I use this basis to reduce matrix $$M$$ on $$\textrm{span}(v_1)^\perp$$? Theoretically, the reduced matrix should be $$3\times 3$$ and have the remaining three eigenvectors/ eigenvalues. But I do not see how I can do the reduction and how to get the desired $$3\times 3$$ matrix.

I think this boils down to the question how to describe the restriction of a linear map (which is itself a linear map) with matrix representation A as a matrix A‘.

• There’s a probably more convenient basis that you can work with that’s trivially mechanically derived from $v_1$: $\{(2,-1,0,0)^T, (3,0,-1,0)^T, (4,0,0,-1)^T\}$.
– amd
Commented Apr 11, 2020 at 20:51
• Thank you! Now I think I restrict my linear map with matrix representation A to the subspace which has your basis: How do I get my new („reduced“) matrix which represents the restriction? Commented Apr 11, 2020 at 20:52

Basically, you’re going to perform a partial diagonalization of $$M$$.

Let $$\{v_2,\dots,v_n\}$$ be a basis for the orthogonal complement of $$v_1$$ and assemble $$v_1$$ and the other basis vectors into the matrix $$B$$. Then $$B^{-1}MB = \begin{bmatrix}\lambda_1 & \mathbf 0^T \\ \mathbf 0 & M'\end{bmatrix}.$$ The submatrix $$M'$$ is the “reduced” matrix that you’re looking for. Its eigenvalues are the remaining eigenvalues of $$M$$, but keep in mind that you’re no longer working in the standard basis, so once you’ve found the coordinates of its eigenvectors, you’ll need to convert back into the original basis.

As for what to choose as the basis for $$v_1^\perp$$, I’m not sure that an orthonormal basis is the best practical choice. Because of all the normalization the entries of $$M'$$ are unlikely to be “nice,” which will just make it more likely that you’ll make errors if you’re doing this by hand. An orthogonal basis might be convenient because it makes the inversion for the change of basis easier: $$B^{-1}MB = \frac1{\det B}B^TMB$$, but you have to trade that off against the work required to produce this basis. A basis that’s very simple to generate is $$\{(2,-1,0,0,\dots)^T, (3,0,-1,0,\dots)^T, (4,0,0,-1,\dots)^T, \dots\}$$ and inverting the resulting matrix doesn’t seem like it would be too bad. You could, for instance, perform Gaussian elimination on $$[B\mid MB]$$ to compute $$B^{-1}MB$$.

Using your basis, we compute $$B^{-1}MB = \begin{bmatrix}60&0&0&0\\0&-16&28&-32\\0&-84&-168&102\\0&92&4&-86\end{bmatrix}.$$ The upper-left element is indeed the eigenvalue associated with $$(1,2,3,4)^T$$ and the rest of that row and column consists of zeros, as expected.

• Up to my computations, the reduced matrix is $M':=\begin{pmatrix}-16 & 28 & \color{red}{-32}\\-84 & -168 & 102\\92 & 4 & -86\end{pmatrix}$ and it has the desired eigenvalues $\lambda_2,\lambda_3,\lambda_4$. --- What do you mean with: "[...] but keep in mind that you’re no longer working in the standard basis, so once you’ve found the coordinates of its eigenvectors, you’ll need to convert back into the original basis."? What do I have to transform now? Commented Apr 12, 2020 at 14:49
• @M.Arianne Transcription error on my part. Good catch!. You’ve applied a change of basis to $M$, so any eigenvectors that you find after that are going to be expressed relative to that basis. For instance, $(1,0,0,0)^T$ is obviously an eigenvector of $B^{-1}MB$, but just as obviously, not of the original matrix $M$. That is the coordinate vector of $v_1$ relative to the $B$-basis, though.
– amd
Commented Apr 12, 2020 at 23:23
• I would like to argue that the reduced matrix can have negative eigenvalues only. Is there such an argument? Or is there a special basis of $(\textrm{span}(v_1))^\perp$ for which this can be read off directly, when looking at $B^{T}MB$? Commented Apr 14, 2020 at 15:05
• @M.Arianne Sure, there is—it’s a basis of eigenvectors, but that isn’t the short-cut that you’re looking for. I don’t know of any easy way to tell that all of the eigenvalues of a non-symmetric matrix are negative by examining the matrix itself. There are, however, methods for counting negative roots of a polynomial, so you could compute the characteristic polynomial of $M'$ and apply those. Descartes’ rule of signs, for instance, tells us that $M'$ has no positive eigenvalues and either $3$ or $1$ negative ones.
– amd
Commented Apr 14, 2020 at 19:16
• @M.Arianne Of course, you could also apply these root-counting methods to the characteristic polynomial of $M$ as well.
– amd
Commented Apr 14, 2020 at 19:21