How is the following decomposition done?

Question

\begin{align*} \begin{bmatrix} 1-a & a \\[5pt] b & 1-b \end{bmatrix} &= \frac{1}{a+b} \begin{bmatrix} b & a \\[5pt] b & a \end{bmatrix}+\frac{1-a-b}{a+b} \begin{bmatrix} a & -a \\[5pt] -b & b \end{bmatrix}. \end{align*}

How is the above decomposition done?

Can anyone explain?

Answer 1

Note that $$\frac {1-a-b}{a+b}=\frac {1}{a+b}-1$$ Multiply by \begin{bmatrix} a & -a \\[5pt] -b & b \end{bmatrix} and substitute the result in the RHS to get the LHS.

Note that the LHS is $$I-\begin{bmatrix} a & -a \\[5pt] -b & b \end{bmatrix}$$

Thus you get the same expression on both sides of

$$\begin{align*} \begin{bmatrix} 1-a & a \\[5pt] b & 1-b \end{bmatrix} &= \frac{1}{a+b} \begin{bmatrix} b & a \\[5pt] b & a \end{bmatrix}+\frac{1-a-b}{a+b} \begin{bmatrix} a & -a \\[5pt] -b & b \end{bmatrix}. \end{align*}$$

Answer 2

If you are asking how to see if the equality is true, the answer is do a bit of algebra, and simplify the right hand side. If you're asking how this decomposition was arrived at, I hope my answer will help.

From your cited source, we see that the left hand matrix is called $P$ and we are interested in powers of $P$.

Thus this suggests that the decomposition arises from considering the eigenvalues.

Let $$M=\newcommand\bmat{\begin{pmatrix}}\newcommand\emat{\end{pmatrix}}\bmat a & -a \\ - b & b \emat.$$

The characteristic polynomial is $(t-a)(t-b)-ab = t^2 - (a+b)t=t(t-(a+b))$. Thus the eigenvalues of $M$ are $0$ and $a+b$. (These are not equal, since the source assumes $0 < a+b < 2$, so $M$ is diagonalizable).

Since $P=1-M$, the eigenvalues of $P$ are $1$ and $1-a-b$. These give the coefficients on the right hand side.

Now $M$ clearly has $0$-eigenvector $\bmat 1 \\ 1 \emat$, and subtracting $a+b$ from $M$, we get $$\bmat -b & -a \\ -b & -a \emat,$$ and a vector in the kernel here is $\bmat a \\ -b\emat.$

Thus we end up with the eigenbasis $v=\bmat 1 \\ 1 \emat,$ $w=\bmat a \\ -b \emat.$ Now choose orthogonal vectors for each of these, like $v'=\bmat 1 \\ -1 \emat$ and $w'=\bmat b \\ a \emat.$

Then consider how the matrix $$wv'^T = \bmat a & -a \\ -b & b \emat$$ acts on vectors $xv+yw$, it gives us $$wv'^T(xv+yw)=xwv'^tv+ywv'^Tw=y(a+b)w.$$

Similarly $$vw'^T=\bmat b & a \\ b & a \emat $$ maps $xv+yw$ to $x(a+b)v$.

Then $P$ maps $xv+yw$ to $xv + y(1-a-b)w$, so we find that $P$ and the linear combination $$ \frac{1}{a+b} vw'^T + \frac{1-a-b}{a+b} wv'^T$$ do the same thing to all vectors. Thus we get the desired equality, $$ \bmat 1-a & a \\ b & 1-b \emat = \frac{1}{a+b}\bmat b & a \\ b & a \emat +\frac{1-a-b}{a+b}\bmat a & -a \\ -b & b \emat . $$

Answer 3

To supplement jgon’s answer, this is a specific instance of a more general decomposition of a diagonalizable matrix into a linear combination of projectors onto its eigenspaces. In particular, a $2\times2$ matrix $A$ with distinct eigenvalues $\lambda_1$ and $\lambda_2$ can be decomposed into $\lambda_1P_1+\lambda_2P_2$, with $P_i^2=P_i$, $P_1P_2=P_2P_1=0$ and the image of $P_i$ equal to the eigenspace of $\lambda_i$. The first two properties in turn imply that $$A^n = (\lambda_1P_1+\lambda_2P_2)^n = \lambda_1^nP_1+\lambda_2^nP_2$$ (expand via the binomial theorem and use the above properties of $P_1$ and $P_2$ to simplify—every mixed term vanishes). It’s a useful exercise to work out what $P_1$ and $P_2$ are, but here I’ll simply state without proof that $$P_1 = {A-\lambda_2 I \over \lambda_1-\lambda_2} \\ P_2 = {A-\lambda_1 I \over \lambda_2-\lambda_1}.$$

In your specific problem, the eigenvalues of $P$ can be found by inspection: all of the rows sum to $1$, therefore $\lambda_1=1$ is an eigenvalue (as it always is for a stochastic matrix) and the other, $\lambda_2=1-a-b$, is easily found by using the fact that the sum of the eigenvalues is equal to the trace. We therefore have $$P_1 = {1\over a+b}\begin{bmatrix}b&a\\b&a\end{bmatrix} \\ P_2 = {1\over a+b}\begin{bmatrix}a&-a\\-b&b\end{bmatrix},$$ and so $$P = \lambda_1P_1+\lambda_2P_2 = {1\over a+b}\begin{bmatrix}b&a\\b&a\end{bmatrix} + {1-a-b\over a+b}\begin{bmatrix}a&-a\\-b&b\end{bmatrix}.$$ Observe that the columns of $P_1$ are eigenvectors of $1$ and the columns of $P_2$ are eigenvectors of $1-a-b$.