Find the inverse of the following matrix.

Question

How can I calculate the inverse of $M$ such that:

$M \in M_{2n}(\mathbb{C})$ and $M = \begin{pmatrix} I_n&iI_n \\iI_n&I_n \end{pmatrix}$, and I find that $\det M = 2^n$. I tried to find the $comM$ and apply $M^{-1} = \frac{1}{2^n} (comM)^T$ but I think it's too complicated.

Answer 1

Hint: Compute the inverse of $$\begin{pmatrix}1 & i \\ i & 1\end{pmatrix}$$ and try the same pattern.

Answer 2

$M^{-1} = \begin{pmatrix} 0.5I_n&-0.5iI_n \\-0.5iI_n&0.5I_n \end{pmatrix}$

Answer 3

A general strategy in such cases is to first compute the result for some small values of $n$, then try to discern a general pattern for the answer, formulate your guess, and attempt to prove it.

Answer 4

General block form inverse with $A$ and $D$ invertible is

$$\begin{pmatrix}A & B \\ C & D\end{pmatrix}^{-1}= \begin{pmatrix}X & -A^{-1}BY \\ -D^{-1}CX & Y\end{pmatrix}$$

where $X=(A-BD^{-1}C)^{-1}$ and $Y = (D-CA^{-1}B)^{-1}$

substituting values will give you $X=Y=\frac12I$ and off-diagonals $-\frac12iI$. Putting all together $M^{-1} = \frac12M^*$.

($*$ is for Hermetian transpose)

Answer 5

I like the @gerw's idea. But if you want a direct way, you can use the method below:

If $$M^{-1}=\begin{pmatrix}A & B \\ C & D\end{pmatrix}$$ then, $$\begin{pmatrix}I_{n} & O \\ O & I_{n}\end{pmatrix}=I_{2n}=MM^{-1}=\begin{pmatrix} I_n&iI_n \\iI_n&I_n \end{pmatrix} \begin{pmatrix}A & B \\ C & D\end{pmatrix}=\begin{pmatrix}A+iC & B+iD \\ iA+C & iB+D\end{pmatrix}$$ now you have a system of 4 equation that gives you $$A=D=\frac12I,$$ and $$B=C=-\frac12iI.$$

Answer 6

Yet another way to do it:

Observe that

$M = I + J, \tag{1}$

where

$I = \begin{bmatrix} I_n & 0 \\ 0 & I_n \end{bmatrix} \tag{2}$

and

$J = iP, \tag{2}$

with

$P = \begin{bmatrix} 0 & I_n \\ I_n & 0 \end{bmatrix}. \tag{3}$

Then

$P^2 = I, \tag{4}$

so that

$J^2 = -I, \tag{5}$

from which we have

$(I + J)(I - J) = I^2 - J^2 = I + I = 2I \tag{6}$

and so

$M(I - J) / 2 = (I + J)(I - J) / 2 = I, \tag{7}$

showing that

$M^{-1} = \dfrac{1}{2}(I - J) = \dfrac{1}{2} \begin{bmatrix} I_n & -iI_n \\ -iI_n & I_n \end{bmatrix}. \tag{8}$

The above calculation works for the same readon as, is inspired by, and is a generalization of the ordinary formula

$z^{-1} = \bar z / \vert z \vert^2, \tag{9}$

which holds for any complex number $z$. Indeed, it is easy to see that

$(aI + bJ)^{-1} = (aI - bJ) / (a^2 + b^2) \tag{10}$

by the same general method. For the sake of closure and completeness, note that

$M^{-1} = \dfrac {1}{2}(I - J) = \dfrac{1}{2} M^\dagger, \tag{11}$

where $M^\dagger$ is the conjugate transpose, that is, the Hermitian adjoint, of $M$.

Hope this helps. Cheers,

and as always,

Fiat Lux!!!

Answer 7

You can easily just do gaussian elimination to compute this inverse as you would do for any other explicitly given matrix.

We will use row operations and track what we did on the right hand side, so we start with $$ \pmatrix{I_n & iI_n \\ iI_n & I_n} \,\Bigg|\, \pmatrix{I_n & 0 \\ 0 & I_n}. $$ Multiply the first $n$ rows by $i$ and substract them from the rows $n+1$ to $2n$ and you get $$ \pmatrix{I_n & iI_n \\ 0 & 2I_n} \,\Bigg|\, \pmatrix{I_n & 0 \\ -iI_n & I_n}. $$ Multiply the rows $n+1$ to $2n$ by $\frac 1 2$, $$ \pmatrix{I_n & iI_n \\ \phantom{\frac 1 2}\!\!\!0 & I_n} \,\Bigg|\, \pmatrix{I_n & 0 \\ -\frac 1 2iI_n & \frac 1 2 I_n}. $$ Now multiply the rows $n+1$ to $2n$ by $i$ and substract from the first $n$ rows to get $$ \pmatrix{\phantom{\frac 1 2}\!\!\!I_n & 0 \\ \phantom{\frac 1 2}\!\!\!0 & I_n} \,\Bigg|\, \pmatrix{\frac 1 2 I_n & -\frac 1 2 i I_n \\ -\frac 1 2iI_n & \frac 1 2 I_n}. $$ Thus, $$ \pmatrix{I_n & iI_n \\ iI_n & I_n}^{-1} = \pmatrix{\frac 1 2 I_n & -\frac 1 2 i I_n \\ -\frac 1 2iI_n & \frac 1 2 I_n}. $$

For the determinant we can use row operations as well: $$ \det\pmatrix{I_n & iI_n \\ iI_n & I_n} = \det\pmatrix{I_n & iI_n \\ 0 & 2I_n} = 2^n. $$

Answer 8

Asking for the inverse of such a block matrix is the same as asking the inverse of a $2\times2$ matrix over the non-commutative ring $M_n(\Bbb C)$ (viewing the individual blocks as elements of that ring). This point of view is not really helpful unless some special circumstance arises; in this case those four "entries" all commute with each other (obviously, since they are all multiples of the identity; however the technique that follows applies whenever they commute, taking $R$ to be the ring generated by the entries).

Now a $2\times2$ matrix $(\begin{smallmatrix}A&B\\C&D\end{smallmatrix})$ over any commutative ring$~R$ (and we can see our matrix as such) is invertible if and only if the determinant $\Delta=AD-BC$ is invertible in$~R$, in which case the inverse is given by the usual formula $\Delta^{-1}(\begin{smallmatrix}D&-B\\-C&A\end{smallmatrix})$. In the current case $\Delta=2I_n$ is certainly invertible, so this applies; the computation now is trivial.