Question

Show that if matrices \(F\) and \(G\) can be diagonalized by the same \(C\) matrix, then they commute. Hint: Do diagonal matrices commute?

Short answer

Matrices \(F\) and \(G\) commute because their diagonal forms \(D_F\) and \(D_G\) commute.

Step-by-step solution

- Assume Diagonalizable Matrices

Assume matrices \(F\) and \(G\) are diagonalizable by the same matrix \(C\). This means there exist diagonal matrices \(D_F\) and \(D_G\) such that \(F = C D_F C^{-1}\) and \(G = C D_G C^{-1}\).

- Write the Product FG

Substitute the diagonalizable forms of \(F\) and \(G\) into the product \(FG\). This gives \(FG = (C D_F C^{-1})(C D_G C^{-1})\).

- Simplify the Product FG

Simplify the expression. Using the property of matrix multiplication: \((AB)(CD) = A(BC)D\), we get \(FG = C D_F (C^{-1} C) D_G C^{-1}\). Since \(C^{-1} C\) is the identity matrix, the expression simplifies to \(FG = C D_F D_G C^{-1}\).

- Write the Product GF

Similarly, calculate the product \(GF\). This gives \(GF = (C D_G C^{-1})(C D_F C^{-1})\).

- Simplify the Product GF

Simplify \(GF\) using the same steps as before. We get \(GF = C D_G (C^{-1} C) D_F C^{-1}\), which simplifies to \(GF = C D_G D_F C^{-1}\).

- Compare FG and GF

From the simplifications, \(FG = C D_F D_G C^{-1}\) and \(GF = C D_G D_F C^{-1}\). Since \(D_F\) and \(D_G\) are diagonal matrices, they commute, i.e., \(D_F D_G = D_G D_F\). Therefore, \(FG = GF\).

Key concepts

Matrix Diagonalization

Matrix diagonalization is a process where a square matrix is converted into a diagonal matrix. This is particularly helpful as diagonal matrices are simpler to work with. To diagonalize a matrix, one typically finds a matrix of eigenvectors and a diagonal matrix of eigenvalues. Here's a breakdown of the process:

• Find all eigenvalues of the matrix.
• Find the corresponding eigenvectors for each eigenvalue.
• Form matrix \(C\) from the eigenvectors.
• Construct the diagonal matrix \(D\) from the eigenvalues.

If matrix \(A\) can be written as \(A = C D C^{-1}\), then \(A\) is said to be diagonalizable. Diagonalization has many applications, such as simplifying matrix computations and solving differential equations.

Matrix Multiplication

Matrix multiplication involves the product of two matrices. To multiply matrices \(A\) and \(B\), the number of columns in \(A\) must equal the number of rows in \(B\). The product of matrices \(A\) and \(B\), denoted as \(AB\), is defined such that each entry \(c_{ij}\) in the resulting matrix is computed as:
\begin{displaymath} c_{ij} = \text{sum of the products of the elements of the i-th row of } A \text{ and the j-th column of } B \text{.} \text{Mathematically:} ewline c_{ij} = \bigg( \text{row}_i(A) \times \text{column}_j(B) \bigg) ewline \text{considering } \bigg(\text{row}_i(A)\bigg) = [a_i1, a_i2, \text{\textellipsis}, a_in] ewline \text{and } \bigg(\text{column}_j(B)\bigg) = [b_1j, b_2j, \text{\textellipsis}, b_nj]
The simplified steps can be:

• Matrix elements are multiplied and summed row-wise and column-wise.
• The size of the resulting matrix is the rows of the first matrix by the columns of the second matrix.
• It is crucial to follow the order strictly as matrix multiplication is not commutative, meaning \(AB eq BA\) generally.

Different properties including associative, distributive, but not commutative nature are important concepts here.

Diagonal Matrices

Diagonal matrices are matrices where non-diagonal elements are zero. A general form of a diagonal matrix \(D\) is:
\begin{displaymath} D = \begin{pmatrix} d_1 & 0 & \text{\textellipsis} & 0 \ 0 & d_2 & \text{\textellipsis} & 0 \ \text{\textellipsis} & \text{\textellipsis} & \text{\textellipsis} & \text{\textellipsis}\ 0 & 0 & \text{\textellipsis} & d_n ewline \text{Properties of Diagonal Matrices include:}

• Simplified multiplication (element-wise).
• Easy to compute powers of, as \(D^k\) results in exponentiation of the diagonal elements.
• Diagonal matrices are commutative when multiplied together: for matrices \(D_1\) and \(D_2\), \(D_1D_2 = D_2D_1\).

This property particularly simplifies the proof in answering the earlier solution as the commutative nature of diagonal matrices directly implies that \(FG=GF\).