Problem 48 - Differential Equations and Linear Algebra

Question

Suppose that \(A\) and \(B\) are invertible matrices. Prove that the block matrix $$\left(\begin{array}{cc} A & 0 \\ 0 & B^{-1} \end{array}\right)$$ is invertible.

Step-by-Step Solution

Answer

In order to prove that the given block matrix \[ M = \left(\begin{array}{cc} A & 0 \\ 0 & B^{-1} \end{array}\right) \] is invertible, we need to find its inverse matrix, N, such that \(MN = NM = I\), where I is the identity matrix. By multiplying matrices M and N, we can find values for the elements in N, resulting in \(N = \left(\begin{array}{cc} A^{-1} & 0 \\ 0 & B \end{array}\right)\). Finally, we verify that \(MN = NM = I\). Therefore, the given matrix M is invertible.

Step 1: Construct the given matrix

The given block matrix is: \[ M = \left(\begin{array}{cc} A & 0 \\ 0 & B^{-1} \end{array}\right) \]

Step 2: Find the inverse of the given matrix

To find the inverse of the matrix M, we can construct a matrix N, the inverse of M, which satisfies the following condition: \[ MN = NM = I, \] where I is the identity matrix. Let the matrix N be: \[ N = \left(\begin{array}{cc} X & Y \\ Z & W \end{array}\right) \] Now, we want to find the matrices X, Y, Z, and W to satisfy the inverse condition.

Step 3: Find the matrices X, Y, Z, and W to satisfy the inverse condition

First, we multiply M and N: \[ \left(\begin{array}{cc} A & 0 \\ 0 & B^{-1} \end{array}\right) \left(\begin{array}{cc} X & Y \\ Z & W \end{array}\right) = \left(\begin{array}{cc} AX + 0Z & AY + 0W \\ 0X + B^{-1}Z & 0Y + B^{-1}W \end{array}\right) = \left(\begin{array}{cc} AX & AY \\ B^{-1}Z & B^{-1}W \end{array}\right) \] Simultaneously, we get the product of N and M: \[ \left(\begin{array}{cc} X & Y \\ Z & W \end{array}\right) \left(\begin{array}{cc} A & 0 \\ 0 & B^{-1} \end{array}\right) = \left(\begin{array}{cc} XA + Y0 & X0 + YB^{-1} \\ ZA + W0 & Z0 + WB^{-1} \end{array}\right) = \left(\begin{array}{cc} XA & YB^{-1} \\ ZA & WB^{-1} \end{array}\right) \] We know that MN = NM = I, so we can write the following equalities: - AX = XA = I - AY = YB^{-1} = 0 - B^{-1}Z = ZA = 0 - B^{-1}W = WB^{-1} = I Since A and B are invertible, we can easily find the following values: - X = A^{-1} - Y = 0 - Z = 0 - W = B

Step 4: Construct the inverse matrix N

Using the values found in Step 3, we can construct the inverse matrix N as follows: \[ N = \left(\begin{array}{cc} A^{-1} & 0 \\ 0 & B \end{array}\right) \]

Step 5: Verify the inverse property

Lastly, we need to verify that MN = NM = I. We do this by multiplying M and N as calculated in Step 3: \[ MN = \left(\begin{array}{cc} A & 0 \\ 0 & B^{-1} \end{array}\right) \left(\begin{array}{cc} A^{-1} & 0 \\ 0 & B \end{array}\right) = \left(\begin{array}{cc} AA^{-1} & 0 \\ 0 & B^{-1}B \end{array}\right) = \left(\begin{array}{cc} I & 0 \\ 0 & I \end{array}\right) = I \] \[ NM = \left(\begin{array}{cc} A^{-1} & 0 \\ 0 & B \end{array}\right) \left(\begin{array}{cc} A & 0 \\ 0 & B^{-1} \end{array}\right) = \left(\begin{array}{cc} A^{-1}A & 0 \\ 0 & BB^{-1} \end{array}\right) = \left(\begin{array}{cc} I & 0 \\ 0 & I \end{array}\right) = I \] Since MN = NM = I, the given matrix M is invertible with inverse matrix N.

Key Concepts

Block Matrices

Block matrices are a special way to visualize and manipulate large matrices by dividing them into smaller sub-matrices or blocks. This decomposition makes dealing with complex matrices more manageable. In particular, block matrices can often simplify the process of computing matrix inversions, which can otherwise be computationally prohibitive due to their larger size.
Block matrices are represented like this:

• Each block can be a matrix in its own right.
• Operations on block matrices often rely on operations on the individual blocks.

When dealing with block matrices like:\[\left(\begin{array}{cc} A & B \ C & D \end{array}\right)\]it's essential to understand how each sub-matrix interacts to perform computations like inversion. By focusing on each block separately, you can simplify and tailor your approach to tackle individual parts of the matrix while keeping the overall structure in context.

Invertible Matrices

An invertible matrix, also known as a nonsingular or non-degenerate matrix, is one that has an inverse. The inverse of a matrix \(A\) is another matrix \(A^{-1}\) satisfying the equation:

• \(AA^{-1} = A^{-1}A = I\)

where \(I\) is the identity matrix, which is a square matrix with 1s on its diagonal and 0s elsewhere, effectively acting as the matrix equivalent of the number 1 in multiplication.
Not all matrices are invertible. For a matrix to be invertible:

• It must be square (having the same number of rows and columns).
• Its determinant must be non-zero.

Invertible matrices are critical in solving systems of linear equations, transforming coordinate systems, and numerous other applications. When a block matrix is composed of invertible sub-matrices, like \(A\) and \(B^{-1}\) in our exercise, each of these blocks must individually satisfy the invertibility conditions for the entire block matrix to remain invertible.

Matrix Multiplication

Matrix multiplication is a fundamental operation where two matrices are combined to produce a third matrix. It's not as straightforward as multiplying individual numbers and has strict rules:

• The number of columns in the first matrix must match the number of rows in the second matrix.
• The resulting matrix will have the dimensions of the outer dimensions of the two matrices being multiplied.

The multiplication is done by taking the row elements of the first matrix and column elements of the second matrix, multiplying them element-wise, and summing the result to form a new element.
Given the matrices:\[A = \left(\begin{array}{cc} a_{11} & a_{12} \ a_{21} & a_{22} \end{array}\right), B = \left(\begin{array}{cc} b_{11} & b_{12} \ b_{21} & b_{22} \end{array}\right)\]The resulting matrix \(C = AB\) will be: \[C = \left(\begin{array}{cc}a_{11}b_{11} + a_{12}b_{21} & a_{11}b_{12} + a_{12}b_{22} \ a_{21}b_{11} + a_{22}b_{21} & a_{21}b_{12} + a_{22}b_{22}\end{array}\right)\]
Understanding matrix multiplication is crucial for finding inverses of block matrices. In the exercise given, matrix multiplication is used to verify that the proposed inverse of the block matrix indeed yields the identity matrix, hence proving the block matrix is invertible.