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stellarik [79]
3 years ago
7

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

Mathematics
1 answer:
Romashka [77]3 years ago
3 0

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.

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