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Matrices - Transpose of a matrix, symmetric and skew symmetric matrices

Grade 12CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Transpose of a Matrix: The transpose of a matrix AA (denoted as ATA^T or A′A') is created by interchanging its rows and columns. If the original matrix AA is of order m×nm \times n, its transpose ATA^T will be of order n×mn \times m. Visually, this operation corresponds to a reflection or 'flip' of the matrix elements across the main diagonal, which runs from the top-left to the bottom-right corner.

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Properties of Transpose: There are four fundamental properties: 1) The transpose of a transpose returns the original matrix: (AT)T=A(A^T)^T = A. 2) The transpose of a sum is the sum of transposes: (A+B)T=AT+BT(A + B)^T = A^T + B^T. 3) A scalar kk remains unchanged: (kA)T=kAT(kA)^T = k A^T. 4) The Reversal Law for multiplication: (AB)T=BTAT(AB)^T = B^T A^T, meaning the order of multiplication is reversed when the transpose is applied.

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Symmetric Matrix: A square matrix AA is called symmetric if it is equal to its transpose (AT=AA^T = A). In terms of elements, aij=ajia_{ij} = a_{ji} for all ii and jj. Visually, the matrix appears as a perfect mirror image across the main diagonal, with entries above the diagonal being identical to their corresponding entries below the diagonal.

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Skew-Symmetric Matrix: A square matrix AA is called skew-symmetric if AT=−AA^T = -A. This implies that for all elements, aij=−ajia_{ij} = -a_{ji}. A critical visual and mathematical feature is that all diagonal elements of a skew-symmetric matrix must be zero, because aii=−aiia_{ii} = -a_{ii} leads to 2aii=02a_{ii} = 0, or aii=0a_{ii} = 0. Off-diagonal elements are negatives of their mirrors across the diagonal.

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Decomposition of a Square Matrix: Any square matrix AA can be uniquely expressed as the sum of a symmetric matrix and a skew-symmetric matrix. This is represented visually by breaking the matrix into two parts: A=P+QA = P + Q, where P=12(A+AT)P = \frac{1}{2}(A + A^T) is symmetric and Q=12(A−AT)Q = \frac{1}{2}(A - A^T) is skew-symmetric.

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Symmetry in Matrix Products: For any matrix AA, the product of the matrix and its transpose (either AATAA^T or ATAA^T A) always results in a symmetric matrix. This is a common property used to verify calculations in higher-level linear algebra.

📐Formulae

Transpose: A=[aij]m×n  ⟹  AT=[aji]n×mA = [a_{ij}]_{m \times n} \implies A^T = [a_{ji}]_{n \times m}

Double Transpose Property: (AT)T=A(A^T)^T = A

Sum Property: (A+B)T=AT+BT(A + B)^T = A^T + B^T

Scalar Multiplication: (kA)T=kAT(kA)^T = k A^T

Reversal Law: (AB)T=BTAT(AB)^T = B^T A^T

Symmetric Matrix Condition: AT=A  ⟺  aij=ajiA^T = A \iff a_{ij} = a_{ji}

Skew-Symmetric Matrix Condition: AT=−A  ⟺  aij=−ajiA^T = -A \iff a_{ij} = -a_{ji} and aii=0a_{ii} = 0

Sum Representation: A=12(A+AT)+12(A−AT)A = \frac{1}{2}(A + A^T) + \frac{1}{2}(A - A^T)

💡Examples

Problem 1:

Express the matrix A=[24−68]A = \begin{bmatrix} 2 & 4 \\ -6 & 8 \end{bmatrix} as the sum of a symmetric and a skew-symmetric matrix.

Solution:

  1. First, find ATA^T: AT=[2−648]A^T = \begin{bmatrix} 2 & -6 \\ 4 & 8 \end{bmatrix}.
  2. Let PP be the symmetric part: P=12(A+AT)=12([24−68]+[2−648])=12[4−2−216]=[2−1−18]P = \frac{1}{2}(A + A^T) = \frac{1}{2}\left(\begin{bmatrix} 2 & 4 \\ -6 & 8 \end{bmatrix} + \begin{bmatrix} 2 & -6 \\ 4 & 8 \end{bmatrix}\right) = \frac{1}{2}\begin{bmatrix} 4 & -2 \\ -2 & 16 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ -1 & 8 \end{bmatrix}.
  3. Let QQ be the skew-symmetric part: Q=12(A−AT)=12([24−68]−[2−648])=12[010−100]=[05−50]Q = \frac{1}{2}(A - A^T) = \frac{1}{2}\left(\begin{bmatrix} 2 & 4 \\ -6 & 8 \end{bmatrix} - \begin{bmatrix} 2 & -6 \\ 4 & 8 \end{bmatrix}\right) = \frac{1}{2}\begin{bmatrix} 0 & 10 \\ -10 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 5 \\ -5 & 0 \end{bmatrix}.
  4. Verify A=P+QA = P + Q: [2−1−18]+[05−50]=[24−68]=A\begin{bmatrix} 2 & -1 \\ -1 & 8 \end{bmatrix} + \begin{bmatrix} 0 & 5 \\ -5 & 0 \end{bmatrix} = \begin{bmatrix} 2 & 4 \\ -6 & 8 \end{bmatrix} = A.

Explanation:

The solution utilizes the theorem that any square matrix AA can be decomposed into PP (symmetric) and QQ (skew-symmetric). We compute the sum and difference of AA and ATA^T and multiply by 12\frac{1}{2} to find these parts.

Problem 2:

If AA and BB are symmetric matrices of the same order, prove that AB−BAAB - BA is a skew-symmetric matrix.

Solution:

  1. Given that AA and BB are symmetric, we know AT=AA^T = A and BT=BB^T = B.
  2. Let X=AB−BAX = AB - BA. To prove XX is skew-symmetric, we must show XT=−XX^T = -X.
  3. Calculate XTX^T: XT=(AB−BA)TX^T = (AB - BA)^T.
  4. Use the property (M−N)T=MT−NT(M - N)^T = M^T - N^T: XT=(AB)T−(BA)TX^T = (AB)^T - (BA)^T.
  5. Apply the Reversal Law (MN)T=NTMT(MN)^T = N^T M^T: XT=(BTAT)−(ATBT)X^T = (B^T A^T) - (A^T B^T).
  6. Substitute the given AT=AA^T = A and BT=BB^T = B: XT=BA−ABX^T = BA - AB.
  7. Factor out −1-1: XT=−(AB−BA)=−XX^T = -(AB - BA) = -X.
  8. Since XT=−XX^T = -X, the matrix AB−BAAB - BA is skew-symmetric.

Explanation:

This proof relies on the properties of matrix transposition and the definitions of symmetric and skew-symmetric matrices. By applying the reversal law and the distributive property of the transpose, we demonstrate the skew-symmetric identity.