# Finding inverse of matrix using its trace

Is there a way to find the inverse of a matrix using its trace?

Maybe by using "Eigendecomposition"(sometimes called "Spectral Decomposition") of a matrix? If it is given that it has an inverse and can be eigendecomposed.

Here is what I am trying to do.

1. Say I have a matrix $A$, which can be eigendecomposed and if none of its eigenvalues are zero, then $A$ is nonsingular and its inverse is given by

${ A }^{ -1 }=Q{ \Lambda }^{ -1 }{ Q }^{ -1 }$

where $Q$ is the square ($N\times N$) matrix whose $i$-th column is the eigenvector $q_{i}$ of $A$ and $\Lambda$ is the diagonal matrix whose diagonal elements are the corresponding eigenvalues, i.e., $\Lambda _{ii}=\lambda _{i}$.

because $\Lambda$ is a diagonal matrix, we can easily calculate its inverse as:

$\left[ { \Lambda }^{ -1 } \right] { ii }=\frac { 1 }{ { \lambda }{ i } }$

2. Now I could use the trace and the determinant of the given matrix to find the eigenvalues. Plug that in the above equation and that's it.

But is there a way to unify both the above properties to get an equation or a system of equations which only requires the trace of the matrix?

• Isn't the answer trivially no? I mean, for any value of the trace you can find infinite matrices of the same size which have that trace, some of which are singular, and some of which are invertible. So from the trace you cannot determine which of those infinite inverses is "the right one." – feralin Dec 28 '15 at 7:20
• Maybe by using Jordan canonical form? – hxri Dec 28 '15 at 7:27
• If you have the Jordan canonical form, then you don't have only the trace. Can you explain what you mean? – feralin Dec 28 '15 at 7:29
• “Now i could use the trace and the determinant of the given matrix to find the eigenvalues.” I am not seeing how this is supposed to work. The two $3x3$ diagonal matrices $\mathrm{diag}(3,-1,6)$ and $\mathrm{diag}(9,-2,1)$ but are very different (the same goes for their inverses). – Jendrik Stelzner Dec 28 '15 at 7:34
• @feralin Well what i mean is that, when you mentioned that the answer is trivial and how can we determine which of those infinite matrices is "the right one", i thought i could use the characteristic polynomial of the matrix and use it to find the similar matrix, and say ${ A }^{ -1 }$ is similar to ${ A }$ over $\mathbb{F}$ (the underlying field) – hxri Dec 28 '15 at 7:43

Firstly, just because $A$ is nonsingular does not mean that it is diagonalizable. For example, $$A=\begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix}$$ is nonsingular but also not diagonalizable. So your first item is not always applicable. Instead, what we can say is that any matrix $A$ (singular or not) can be put into Jordan canonical form. So, typically, it will take more than just the trace and the determinant to find the eigenvalues (or the inverse) of a given matrix. The problem is that for an $n\times n$ matrix, the determinant and the trace are just two conditions on the $n$ (assuming that we are working over an algebraically closed field) eigenvalues, so we have an underdetermined system of equations: we can't really say much about the solution, i.e. the eigenvalues. For example, the matrix $A$ above is nonsingular, has trace $2$ and determinant $1$, but so does the $2\times 2$ identity matrix, and they clearly have different inverses: the inverse of the identity matrix is itself, and the inverse of $A$ is given by $$A^{-1}=\begin{bmatrix} 1 & -1\\ 0 & 1\end{bmatrix}$$
Think of finding the inverse as needing somewhere on the order of $n^3$ operations (though why it is around $n^3$ or, more accurately, $n^{2.373}$ is pretty technical and takes a fair amount of work). Knowing the trace of a matrix simply isn't enough to tell you what the inverse of it is. You only know $1$ thing, and you need to know more than that.