Name: Introduction to Inverse of Matrices
Uploaded: 2019-07-19
Duration: 11 min 27 s
Description: In this video, we will study the inverse of matrices and conditions for the existence of inverses and invertible matrices, along with some of their properties.

Question

Prove that every square matrix is uniquely expressible as the sum of a symmetric matrix and a skew-symmetric matrix.

Accepted Answer

Let $A$ be a square matrix. Then,

$A = \frac{1}{2} (A + A^{T}) + \frac{1}{2} (A - A^{T}) = P + Q$ (say), where $P = \frac{1}{2} (A + A^{T}), Q = \frac{1}{2} (A - A^{T})$ .

Now, $P^{T} = {(\frac{1}{2} (A + A^{T}))}^{T} = \frac{1}{2} {(A + A^{T})}^{T} [∵ (k A)^{T} = k A^{T}]$

$\Rightarrow P^{T} = \frac{1}{2} (A^{T} + {(A^{T})}^{T}) [∵ (A + B)^{T} = A^{T} + B^{T}]$

$\Rightarrow P^{T} = \frac{1}{2} (A^{T} + A) [∵ {(A^{T})}^{T} = A]$

$\Rightarrow P^{T} = \frac{1}{2} (A + A^{T}) = P$ [By commutativity of matrix addition.]

$∴ P$ is a symmetric matrix.

Also, $Q^{T} = {(\frac{1}{2} (A - A^{T}))}^{T} = \frac{1}{2} {(A - A^{T})}^{T} = \frac{1}{2} (A^{T} - {(A^{T})}^{T})$

$\Rightarrow Q^{T} = \frac{1}{2} (A^{T} - A) = - \frac{1}{2} (A - A^{T}) = - Q$

$∴ Q$ is skew-symmetric matrix.

Thus, $A = P + Q$ , where $P$ is a symmetric matrix and $Q$ is a skew-symmetric matrix.

Hence, $A$ is expressible as the sum of a symmetric matrix and a skew-symmetric matrix.

Uniqueness: If possible, let $A = R + S$ , where $R$ is symmetric matrix and $S$ is skew-symmetric matrix. Then,

$A^{T} = (R + S)^{T} = R^{T} + S^{T}$

$\Rightarrow A^{T} = R - S [∵ R^{T} = R, S^{T} = - S]$

Now, $A = R + S$ and $A^{T} = R - S$

$\Rightarrow R = \frac{1}{2} (A + A^{T}) = P, S = \frac{1}{2} (A - A^{T}) = Q$ .

Hence, $A$ is uniquely expressible as the sum of a symmetric matrix and a skew-symmetric matrix.

Prove that every square matrix is uniquely expressible as the sum of a symmetric matrix and a skew-symmetric matrix.

Important Questions on Matrices

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