Suppose that an eigenvalue of matrix A is zero. Prove that A must therefore be singular.
1 answer:
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]](https://tex.z-dn.net/?f=Let%20A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da1%26a2%26a3%5C%5Ca4%26a5%26a6%5C%5Ca7%26a8%26a9%5Cend%7Barray%7D%5Cright%5D)
![Identity Matrix = I = \left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]](https://tex.z-dn.net/?f=Identity%20Matrix%20%3D%20I%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%260%5C%5C0%261%260%5C%5C0%260%261%5Cend%7Barray%7D%5Cright%5D)
.
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]](https://tex.z-dn.net/?f=0%2A%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%260%5C%5C0%261%260%5C%5C0%260%261%5Cend%7Barray%7D%5Cright%5D%20%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%260%260%5C%5C0%260%260%5C%5C0%260%260%5Cend%7Barray%7D%5Cright%5D)
So,
det (A-0.I) = 0 implies that 0 is eigenvalue of matrix A.
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