Question

Suppose that an eigenvalue of matrix A is zero. Prove that A must therefore be singular.

Answer 1

Eigenvalue:

In a linear system of equations, eigenvalues are actually special set of scalars which are associated with these equations.

Singular Matrix:

A matrix whose determinant is Zero is called singular matrix.

Step-by-step explanation:

$Let A = \left[\begin{array}{ccc}a1&a2&a3\\a4&a5&a6\\a7&a8&a9\end{array}\right]$

$Identity Matrix = I = \left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]$

.

If Matrix A is singular it means that

det (A) = 0

det (A-0.I)=0

because $0*\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] = \left[\begin{array}{ccc}0&0&0\\0&0&0\\0&0&0\end{array}\right]$

So,

det (A-0.I) = 0 implies that 0 is eigenvalue of matrix A.