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Matrices - Symmetric and Skew Symmetric Matrices

Grade 12CBSE

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

🔑Concepts

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A square matrix A=[aij]A = [a_{ij}] is called a symmetric matrix if AT=AA^T = A. This means the element at the ii-th row and jj-th column is equal to the element at the jj-th row and ii-th column, i.e., aij=ajia_{ij} = a_{ji} for all i,ji, j.

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A square matrix A=[aij]A = [a_{ij}] is called a skew-symmetric matrix if AT=−AA^T = -A. This means aij=−ajia_{ij} = -a_{ji} for all i,ji, j.

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In a skew-symmetric matrix, all diagonal elements are zero. Since aii=−aiia_{ii} = -a_{ii} for diagonal elements, it follows that 2aii=02a_{ii} = 0, hence aii=0a_{ii} = 0.

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For any square matrix AA with real number entries, A+ATA + A^T is always a symmetric matrix and A−ATA - A^T is always a skew-symmetric matrix.

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Any square matrix AA can be uniquely expressed as the sum of a symmetric matrix and a skew-symmetric matrix using the relation A=12(A+AT)+12(A−AT)A = \frac{1}{2}(A + A^T) + \frac{1}{2}(A - A^T).

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If AA and BB are symmetric matrices of the same order, then AB+BAAB + BA is symmetric, and AB−BAAB - BA is skew-symmetric.

📐Formulae

AT=A  ⟺  SymmetricA^T = A \iff \text{Symmetric}

AT=−A  ⟺  Skew-SymmetricA^T = -A \iff \text{Skew-Symmetric}

A=P+Q where P=12(A+AT) and Q=12(A−AT)A = P + Q \text{ where } P = \frac{1}{2}(A + A^T) \text{ and } Q = \frac{1}{2}(A - A^T)

aii=0 for all i in a skew-symmetric matrixa_{ii} = 0 \text{ for all } i \text{ in a skew-symmetric matrix}

(A+B)T=AT+BT(A + B)^T = A^T + B^T

(kA)T=kAT(kA)^T = kA^T

💡Examples

Problem 1:

Express the matrix A=[351−1]A = \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix} as the sum of a symmetric and a skew-symmetric matrix.

Solution:

Step 1: Find ATA^T. AT=A^T = \begin{bmatrix} 3 & 1 \ 5 & -1 \end{bmatrix} Step 2: Calculate the symmetric part $P = \frac{1}{2}(A + A^T)$. P = \frac{1}{2} \left( \begin{bmatrix} 3 & 5 \ 1 & -1 \end{bmatrix}++\begin{bmatrix} 3 & 1 \ 5 & -1 \end{bmatrix}\right) = \frac{1}{2}\begin{bmatrix} 6 & 6 \ 6 & -2 \end{bmatrix}==\begin{bmatrix} 3 & 3 \ 3 & -1 \end{bmatrix} Step 3: Calculate the skew-symmetric part $Q = \frac{1}{2}(A - A^T)$. Q = \frac{1}{2} \left( \begin{bmatrix} 3 & 5 \ 1 & -1 \end{bmatrix}−-\begin{bmatrix} 3 & 1 \ 5 & -1 \end{bmatrix}\right) = \frac{1}{2}\begin{bmatrix} 0 & 4 \ -4 & 0 \end{bmatrix}==\begin{bmatrix} 0 & 2 \ -2 & 0 \end{bmatrix} Step 4: Verify $A = P + Q$. P + Q = \begin{bmatrix} 3 & 3 \ 3 & -1 \end{bmatrix}++\begin{bmatrix} 0 & 2 \ -2 & 0 \end{bmatrix}==\begin{bmatrix} 3 & 5 \ 1 & -1 \end{bmatrix}=A = A

Explanation:

We use the theorem that every square matrix can be written as A=P+QA = P + Q, where PP is symmetric (PT=PP^T = P) and QQ is skew-symmetric (QT=−QQ^T = -Q).

Problem 2:

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

Solution:

Let C=AB−BAC = AB - BA. To prove CC is skew-symmetric, we must show CT=−CC^T = -C. Given AT=AA^T = A and BT=BB^T = B. CT=(AB−BA)TC^T = (AB - BA)^T Using the property (X−Y)T=XT−YT(X - Y)^T = X^T - Y^T: CT=(AB)T−(BA)TC^T = (AB)^T - (BA)^T Using the reversal law (XY)T=YTXT(XY)^T = Y^T X^T: CT=BTAT−ATBTC^T = B^T A^T - A^T B^T Substituting AT=AA^T = A and BT=BB^T = B: CT=BA−ABC^T = BA - AB Taking the negative sign common: CT=−(AB−BA)C^T = -(AB - BA) CT=−CC^T = -C

Explanation:

By applying the properties of transposes (distribution over subtraction and the reversal law for multiplication), we demonstrate that the transpose of (AB−BA)(AB - BA) results in its negative, satisfying the definition of a skew-symmetric matrix.