ISRO2008-34

Question

If a square matrix A satisfies $A^TA=I$, then the matrix $A$ is

• A. Idempotent
• B. Symmetric
• C. Orthogonal
• D. Hermitian

Comments

The matrix M is idempotent if and only if MM = M.

A Hermitian matrix (or self-adjoint matrix) is a complex square matrix that is equal to its own conjugate transpose—that is, the element in the i-th row and j-th column is equal to the complex conjugate of the element in the j-th row and i-th column, for all indices i and j:

 or , in matrix form.

Answer 1

If a square matrix A satisfies A^TA=I, then the matrix A is   Orthogonal Matrix.

Option C is correct.

Answer 2

In linear algebra, an orthogonal matrix is a square matrix with real entries whose columns and rows are orthogonal unit vectors (i.e., orthonormal vectors), i.e.

{\displaystyle Q^{\mathrm {T} }Q=QQ^{\mathrm {T} }=I,}

where I is the identity matrix.

https://en.wikipedia.org/wiki/Orthogonal_matrix

Answer 3

Answer is orthogonal matrix.

reference:: http://mathworld.wolfram.com/OrthogonalMatrix.html

Answer 4

Idempotent Matrix: $AA=A^2=A$

Symmetric Matrix: $A=A^T$

Orthogonal Matrix: $AA^T = I$ (Option C)

A Hermitian matrix (or self-adjoint matrix) is a complex square matrix that is equal to its own conjugate transpose.

 

Bonus:-

Singular Matrix: Matrix whose determinant is 0. It's opposite to Invertible matrix, whose determinant is $\neq$ 0

Skew Symmetric Matrix: $A^T=-A$

Comments

Orthogonal Matrix$:A^{T} = A^{-1}$

Multiply both sides with $'A',$ then we get

$AA^{T} = AA^{-1}\implies AA^{T} = I\rightarrow (1)$

On the equation $(1),$ apply the determinant on both sides, then we get

$ \mid AA^{T} \mid = \mid I \mid $

$\implies \mid A \mid \mid A^{T} \mid  = \mid I \mid$

$\implies \mid A \mid ^{2} = 1 \:\:\:\:\: \bigg[\because\: \mid A \mid = \mid A^{T} \mid  and \mid I \mid = 1\bigg]$

$ \implies \mid A \mid  = \pm 1$

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Whoa, thank you (: