Transpose, Symmetric, and Skew-Symmetric Matrices

1. Transpose of a Matrix

The transpose of a matrix A, denoted by $A^T$ or $A'$, is the matrix you get by flipping the original matrix over its main diagonal. This operation effectively interchanges the rows and columns. The first row becomes the first column, the second row becomes the second column, and so on.

• Rule: If $A = [a_{ij}]$ is an m x n matrix, then its transpose $A^T$ will be an n x m matrix where the element at the i-th row and j-th column is the original $a_{ji}$.
• Example: If $A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}$ (a 2x3 matrix), then its transpose $A^T$ is found by making the first row (1, 2, 3) the first column, and the second row (4, 5, 6) the second column: $$A^T = \begin{bmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{bmatrix} \quad (\text{a 3x2 matrix})$$

Properties of Transpose:

For any matrices A and B of suitable orders:

1. $(A^T)^T = A$
   • (Explanation: If you flip the matrix, and then flip it back, you get the original matrix.)
2. $(kA)^T = kA^T$ (where k is any constant)
   • (Explanation: You can either scale the matrix first and then flip it, or flip it first and then scale it. The result is the same.)
3. $(A+B)^T = A^T + B^T$
   • (Explanation: You can either add two matrices and then flip the result, or flip both matrices first and then add them. The result is the same.)
4. $(AB)^T = B^T A^T$ (Reversal Law)
   • (Explanation: This is the most important property. Like taking off socks and shoes, the order of the operation is reversed when transposed. The transpose of a product is the product of the transposes in the reverse order.)

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2. Symmetric Matrix

A square matrix A is called symmetric if it is its own transpose, meaning it is perfectly symmetrical across its main diagonal.

$$A^T = A$$

• Condition: This means that the element in row 'i', column 'j' must be equal to the element in row 'j', column 'i' (i.e., $a_{ij} = a_{ji}$ for all i, j).
• Example: In the matrix below, notice how $a_{12}=a_{21}=7$ and $a_{13}=a_{31}=3$. $$A = \begin{bmatrix} 1 & 7 & 3 \\ 7 & 2 & 5 \\ 3 & 5 & 6 \end{bmatrix} \quad \text{and} \quad A^T = \begin{bmatrix} 1 & 7 & 3 \\ 7 & 2 & 5 \\ 3 & 5 & 6 \end{bmatrix}$$ Since $A = A^T$, this matrix is symmetric.

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3. Skew-Symmetric Matrix

A square matrix A is called skew-symmetric if its transpose is equal to its negative. It is a negative mirror image of itself across the diagonal.

$$A^T = -A$$

• Condition: This means that $a_{ij} = -a_{ji}$ for all i, j.
• Property: This condition has a special consequence for the principal diagonal elements. For any diagonal element $a_{ii}$ (like $a_{11}$ or $a_{22}$), the condition is $a_{ii} = -a_{ii}$. This implies $2a_{ii}=0$, so $a_{ii}=0$. All principal diagonal elements of a skew-symmetric matrix must be zero.
• Example: Notice the diagonal is all zeros, and $a_{12}=1$ while $a_{21}=-1$. $$A = \begin{bmatrix} 0 & 1 & -2 \\ -1 & 0 & 3 \\ 2 & -3 & 0 \end{bmatrix} \quad \text{and} \quad A^T = \begin{bmatrix} 0 & -1 & 2 \\ 1 & 0 & -3 \\ -2 & 3 & 0 \end{bmatrix}$$ You can see that $A^T = -A$, so this matrix is skew-symmetric.

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4. Key Theorems

• Theorem 1: For any square matrix A, the matrix $A+A^T$ is always symmetric, and the matrix $A-A^T$ is always skew-symmetric.

  • (Proof for $A+A^T$: Let's transpose it. $$(A+A^T)^T = A^T + (A^T)^T = A^T + A = A+A^T$$ Since its transpose is equal to itself, it is symmetric.)
  • (Proof for $A-A^T$: Let's transpose it. $$(A-A^T)^T = A^T - (A^T)^T = A^T - A = -(A-A^T)$$ Since its transpose is equal to its negative, it is skew-symmetric.)

• Theorem 2: Any square matrix A can be expressed as the sum of a symmetric and a skew-symmetric matrix. This is a clever algebraic trick. We can write any matrix A as: $$A = \frac{1}{2}(A+A) = \frac{1}{2}(A + A^T + A - A^T)$$ Rearranging this gives:

$$A = \underbrace{\frac{1}{2}(A+A^T)}_{\text{Symmetric Part}} + \underbrace{\frac{1}{2}(A-A^T)}_{\text{Skew-Symmetric Part}}$$

This formula is very useful as it allows us to decompose any square matrix into these two special types of matrices.

Example 1: Finding the Transpose

Question: Find the transpose of the matrix $A = \begin{bmatrix} 3 & 5 \\ 7 & -1 \end{bmatrix}$.

Explanation:

1. Understand Transpose: The transpose of a matrix, $A^T$, is found by interchanging its rows and columns.
2. Identify Rows: The first row of A is [3 5]. The second row of A is [7 -1].
3. Create Columns: The first row of A becomes the first column of $A^T$. The second row of A becomes the second column of $A^T$.
4. Construct the Transpose: $$A^T = \begin{bmatrix} 3 & 7 \\ 5 & -1 \end{bmatrix}$$

Answer: $A^T = \begin{bmatrix} 3 & 7 \\ 5 & -1 \end{bmatrix}$.

Example 2: Verifying $(A+B)^T = A^T+B^T$

Question: If $A = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 3 & 4 \\ 1 & 0 \end{bmatrix}$, verify that $(A+B)^T = A^T+B^T$.

Explanation: 1. Calculate the Left-Hand Side (LHS):

• First, find $A+B$: $$A+B = \begin{bmatrix} 1+3 & 2+4 \\ 0+1 & 1+0 \end{bmatrix} = \begin{bmatrix} 4 & 6 \\ 1 & 1 \end{bmatrix}$$
• Now, find the transpose of the result: $$(A+B)^T = \begin{bmatrix} 4 & 1 \\ 6 & 1 \end{bmatrix}$$

2. Calculate the Right-Hand Side (RHS):

• First, find $A^T$ and $B^T$: $$A^T = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}$$ and $$B^T = \begin{bmatrix} 3 & 1 \\ 4 & 0 \end{bmatrix}$$
• Now, add the transposes: $$A^T+B^T = \begin{bmatrix} 1+3 & 0+1 \\ 2+4 & 1+0 \end{bmatrix} = \begin{bmatrix} 4 & 1 \\ 6 & 1 \end{bmatrix}$$

3. Conclusion:

• Since LHS = $\begin{bmatrix} 4 & 1 \\ 6 & 1 \end{bmatrix}$ and RHS = $\begin{bmatrix} 4 & 1 \\ 6 & 1 \end{bmatrix}$, the property is verified.

Answer: LHS = RHS = $\begin{bmatrix} 4 & 1 \\ 6 & 1 \end{bmatrix}$.

Example 3: Verifying $(AB)^T = B^TA^T$ (Reversal Law)

Question: If $A = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix}$ and $B = \begin{bmatrix} -1 & 2 & 1 \end{bmatrix}$, verify the reversal law for transpose.

Explanation: 1. Calculate the Left-Hand Side (LHS):

• First, find the product $AB$. A is (3x1) and B is (1x3), so $AB$ will be (3x3). $$AB = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix} \begin{bmatrix} -1 & 2 & 1 \end{bmatrix}$$ $$= \begin{bmatrix} 1(-1) & 1(2) & 1(1) \\ -4(-1) & -4(2) & -4(1) \\ 3(-1) & 3(2) & 3(1) \end{bmatrix}$$ $$= \begin{bmatrix} -1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3 \end{bmatrix}$$
• Now, find the transpose of $AB$: $$(AB)^T = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}$$

2. Calculate the Right-Hand Side (RHS):

• First, find $B^T$ and $A^T$: $B^T = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix}$ (a 3x1 matrix).

  $A^T = \begin{bmatrix} 1 & -4 & 3 \end{bmatrix}$ (a 1x3 matrix).

• Now, find the product $B^TA^T$: (3x1) $\times$ (1x3) $\to$ (3x3). $$B^TA^T = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix} \begin{bmatrix} 1 & -4 & 3 \end{bmatrix}$$ $$= \begin{bmatrix} -1(1) & -1(-4) & -1(3) \\ 2(1) & 2(-4) & 2(3) \\ 1(1) & 1(-4) & 1(3) \end{bmatrix}$$ $$= \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}$$

3. Conclusion:

• Since LHS = RHS, the reversal law $(AB)^T = B^TA^T$ is verified.

Answer: LHS = RHS = $\begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}$.

Example 4: Identifying Symmetric and Skew-Symmetric Matrices

Question: Determine if the matrix $A = \begin{bmatrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{bmatrix}$ is symmetric or skew-symmetric.

Explanation:

1. Find the Transpose ($A^T$): We interchange rows and columns.

   • Row 1 [1 -1 5] becomes Column 1.
   • Row 2 [-1 2 1] becomes Column 2.
   • Row 3 [5 1 3] becomes Column 3. $$A^T = \begin{bmatrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{bmatrix}$$

2. Compare $A^T$ with $A$:

   We check if $A^T = A$. By direct comparison, $A^T$ is identical to $A$.

Answer: Since $A^T = A$, the matrix is symmetric.

Example 5: Identifying a Skew-Symmetric Matrix

Question: Show that the matrix $A = \begin{bmatrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{bmatrix}$ is skew-symmetric.

Explanation:

1. Find the Transpose ($A^T$): $$A^T = \begin{bmatrix} 0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0 \end{bmatrix}$$

2. Find the Negative of A ($-A$): $$-A = -1 \times \begin{bmatrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0 \end{bmatrix}$$

3. Compare $A^T$ with $-A$:

   We observe that $A^T$ is identical to $-A$.

Answer: Since $A^T = -A$, the matrix is skew-symmetric.

Example 6: Using Properties of Symmetric/Skew-Symmetric Matrices

Question: If A is a symmetric matrix, show that $B^TAB$ is also symmetric.

Explanation:

1. State the Goal: We need to prove that the matrix $P = B^TAB$ is symmetric. This means we must show that $P^T = P$.

2. State the Given: We are given that A is symmetric, which means $A^T = A$.

3. Take the Transpose: Let's find the transpose of $P$ using the Reversal Law, $$(XY)^T = Y^TX^T$$ We will group $B^TA$ as one matrix and $B$ as the other. $$P^T = ( (B^T A) B )^T = B^T (B^T A)^T$$

4. Apply Reversal Law Again: Now we apply the reversal law to the term $(B^T A)^T$. $$P^T = B^T (A^T (B^T)^T)$$

5. Simplify: We know that $(B^T)^T = B$. So, $$P^T = B^T A^T B$$

6. Use the Given: We were given that A is symmetric, so $A^T = A$. We substitute this into our expression: $P^T = B^T A B$.

7. Conclusion: We have shown that $P^T = B^T A B$, which is equal to the original matrix $P$. Since $P^T = P$, the matrix $B^TAB$ is symmetric.

Answer: Proven.

Example 7: Expressing a Matrix as a Sum

Question: Express the matrix $A = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix}$ as the sum of a symmetric (P) and a skew-symmetric (Q) matrix.

Explanation: We use the formulas $$P = \frac{1}{2}(A+A^T)$$ and $$Q = \frac{1}{2}(A-A^T)$$

1. Find $A^T$: $$A^T = \begin{bmatrix} 2 & -1 & 1 \\ -2 & 3 & -2 \\ -4 & 4 & -3 \end{bmatrix}$$

2. Find the Symmetric Part (P): $$A+A^T = \begin{bmatrix} 2+2 & -2-1 & -4+1 \\ -1-2 & 3+3 & 4-2 \\ 1-4 & -2+4 & -3-3 \end{bmatrix}$$ $$= \begin{bmatrix} 4 & -3 & -3 \\ -3 & 6 & 2 \\ -3 & 2 & -6 \end{bmatrix}$$ $$P = \frac{1}{2}(A+A^T) = \begin{bmatrix} 2 & -3/2 & -3/2 \\ -3/2 & 3 & 1 \\ -3/2 & 1 & -3 \end{bmatrix}$$

3. Find the Skew-Symmetric Part (Q): $$A-A^T = \begin{bmatrix} 2-2 & -2-(-1) & -4-1 \\ -1-(-2) & 3-3 & 4-(-2) \\ 1-(-4) & -2-4 & -3-(-3) \end{bmatrix}$$ $$= \begin{bmatrix} 0 & -1 & -5 \\ 1 & 0 & 6 \\ 5 & -6 & 0 \end{bmatrix}$$ $$Q = \frac{1}{2}(A-A^T) = \begin{bmatrix} 0 & -1/2 & -5/2 \\ 1/2 & 0 & 3 \\ 5/2 & -3 & 0 \end{bmatrix}$$

Answer: $A = P+Q$, where $$P = \begin{bmatrix} 2 & -3/2 & -3/2 \\ -3/2 & 3 & 1 \\ -3/2 & 1 & -3 \end{bmatrix}$$ and $$Q = \begin{bmatrix} 0 & -1/2 & -5/2 \\ 1/2 & 0 & 3 \\ 5/2 & -3 & 0 \end{bmatrix}$$

Example 8: Solving for Variables

Question: If the matrix $A = \begin{bmatrix} 0 & a & 3 \\ 2 & b & -1 \\ c & 1 & 0 \end{bmatrix}$ is a skew-symmetric matrix, find the values of a, b, and c.

Explanation: A skew-symmetric matrix must satisfy two conditions:

1. All diagonal elements are zero: $a_{ii} = 0$.
2. All non-diagonal elements are negative opposites of their counterparts: $a_{ij} = -a_{ji}$.

• Condition 1 (Diagonal): We check the main diagonal: $a_{22} = b$. For the matrix to be skew-symmetric, $b=0$.
• Condition 2 (Non-Diagonal):
  • $a_{12} = -a_{21} \implies a = -(2) \implies a = -2$.
  • $a_{13} = -a_{31} \implies 3 = -(c) \implies c = -3$.
  • (Check $a_{23}$): $a_{23} = -a_{32} \implies -1 = -(1)$, which is true.

Answer: The values are a=-2, b=0, c=-3.

Example 9: Property of $A-A^T$

Question: For any square matrix A, show that $A-A^T$ is a skew-symmetric matrix.

Explanation:

1. Define the Matrix: Let $P = A-A^T$.

2. State the Goal: To prove $P$ is skew-symmetric, we must show that $P^T = -P$.

3. Take the Transpose: $P^T = (A-A^T)^T$.

4. Apply Transpose Properties: Using the property $$(X-Y)^T = X^T-Y^T$$, we get: $$P^T = A^T - (A^T)^T$$

5. Simplify: Using the property $(X^T)^T = X$, we get: $P^T = A^T - A$.

6. Factor and Conclude: Factor out -1: $$P^T = -(A - A^T)$$ Since $P = A-A^T$, we have $P^T = -P$.

Answer: Since its transpose is equal to its negative, $A-A^T$ is skew-symmetric.

Example 10: Property of $AA^T$

Question: Show that for any matrix A, the products $AA^T$ and $A^TA$ are both symmetric matrices.

Explanation: Part 1: Show $AA^T$ is symmetric.

1. Let $P = AA^T$.
2. We need to show $P^T = P$.
3. $P^T = (AA^T)^T$. Using the reversal law $(XY)^T = Y^TX^T$:
4. $P^T = (A^T)^T A^T = A A^T$. (Since $(A^T)^T = A$).
5. We see that $P^T = P$, so $AA^T$ is symmetric.

Part 2: Show $A^TA$ is symmetric.

1. Let $Q = A^TA$.
2. We need to show $Q^T = Q$.
3. $Q^T = (A^TA)^T$. Using the reversal law:
4. $Q^T = A^T (A^T)^T = A^T A$.
5. We see that $Q^T = Q$, so $A^TA$ is symmetric.

Answer: Both $AA^T$ and $A^TA$ are symmetric.