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If I subtract the matrix $A$ from the identity matrix $I$, $I - A$, is there a meaning to the resulting matrix perhaps given some conditions like invertibility or symmetry? For example,

$$ \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 1-a & -b \\ -c & 1-d \end{bmatrix} $$

For future reference, is there a good reference to lookup answers to such questions when I can't derive the answer on my own?

Chewers Jingoist
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  • You might want to be a bit more specific with your question, it seems very open ended. A good resource would be any good linear algebra textbook for these kinds of questions. – wesssg Jun 13 '16 at 01:30
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    @wesssg My apologies for not being more specific but I'm not sure what I'm looking for. I'm a bit confused but the answers so far are certainly shedding light where there was obscurity. Thanks! – Chewers Jingoist Jun 13 '16 at 01:51

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The difference $I - A$ is sometimes used with respect to projection matrices, where $A$ is the projection onto a subspace and $I - A$ is the projection onto the orthogonal complement of that subspace.

cr3
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If there exists $n\in\mathbb{N}$ with $A^{n}=0$, then $I−A$ is invertible (try to prove this yourself before reading why!).

The reason: multiply out the brackets in the expression $$(I−A)(A^{n−1}+A^{n−2}+\cdots+A+I)$$ and see that you get $I$. It follows that $A^{n−1}+A^{n−2}+\cdots+A+I$ is the inverse of $I−A$ in this case.

By the way, a matrix $A$ with $A^{n}=0$ for some $n\in\mathbb{N}$ is called nilpotent.

By the way, the fact presented above can be used to give a proof of the (basic) fact that all eigenvalues of a nilpotent matrix must equal zero. In particular, this means that the determinant and trace of a nilpotent matrix are always $0$.

If $A$ is nilpotent then so is $\frac{1}{\lambda}A$ for any $\lambda\neq0$. Hence $I-\frac{1}{\lambda}A=\frac{1}{\lambda}(\lambda I-A)$ has an inverse. It follows that $\det{(\lambda I-A)}\neq0$ whenever $\lambda\neq0$ and $A$ is nilpotent, so the characteristic polynomial only has $0$ as a root.

So considering $I-A$ can actually come in handy to prove things.

Will R
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If A is invertible the resulting matrix B produced when you subtract $I$ from A may or may not be invertible. Can you think of examples? If A is symmetric we have A = $A^t$, so, we have I - A = $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} a & b \\ b & d \end{bmatrix} = \begin{bmatrix} 1-a & -b \\ -b & 1-d \end{bmatrix}$ is also symmetric. You see something similar to this when you are calculating the characteritic polynomial of a matrix to calculate eigenvectors and test for the diagbalizability of a matrix. but instead you are looking at Det ( A - cI)= 0, where c is in your field.

wesssg
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