A simple system of algebraic (not differential) equations such as:
$$ 3 x + 4 y - 8 z = 23 $$$$ -2 x + 8 y - 11 z = -32 $$$$ -4 x + 9 y - 32 z = - 8 $$
can be written as:$$\mathbf{A} \vec x = \vec y$$
with, of course, the coefficients of $x$, $y$, and $z$ being arranged into a matrix $\mathbf A$ and the 23, -32, and -8 being arranged into a column vector $\vec b$.
My question is: How can we use eigenvectors and eigenvalues to solve for the vector $\vec x$ when the elements of $\mathbf{A}$ and $\vec y$ are known?
A couple of things:
The StackExchange question here somewhat addresses the issue, but the answer given is hard to follow because it goes off on tangents, and addresses far to much for me to find my answer in it. The wikipedia page on eigenvectors and eigenvalues has the same problem.
I am perfectly comfortable finding eigenvalues and eigenvectors, I just don't know how to use them to solve systems of linear equations.
Most google searches that I have tried only show me how to solve systems of linear differential equations.
I am aware that using eigenvalues and eigenvectors is not the usual (or easiest) method of solving systems of algebraic equations.
I am familiar with the general interpretation of eigenvalues and eigenvectors as they pertain to rotating vector spaces, etc.
1 Answer
$\begingroup$In general knowing the eigenvalues and eigenvectors (and generalized eigenvectors if applicable) does not really help you much with finding an exact solution to the linear system $Ax=b$, since it winds up writing the equation as $PJP^{-1}x=b$ where $P$ is the matrix of eigenvectors (and generalized eigenvectors if applicable) and $J$ is the Jordan form. If you were given $P$ and $J$ then you would proceed this way:
- Solve $Py=b$.
- Solve $Jz=y$.
- Take $x=Pz$.
Generally speaking that first step is not any easier than solving $Ax=b$ was in the first place. It is easier if $A$ is normal (i.e. $A^*A = A A^*$ where $A^*$ denotes the conjugate transpose of $A$), because in this case $P$ is unitary so $y=P^* b$, but otherwise there is no particular reason to expect this to be easier. The remaining two steps are cheap by comparison.
When $A$ is not normal, the SVD is more useful for solving the linear system, even when $A$ is not normal, because all three matrices in the factorization are inexpensive to invert. The situation is similar for the QR decomposition.
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