Maths: Linear Algebra

Question

Any shortcut?

Answer 1

Try to solve |T- $\lambda$I | = 0. In first step perform c1 <- c1 + c2 +c3+c4+c5 and after that all elements of first column would be 1+2+3+4+5-$\lambda$ = 15-$\lambda$. Now we take 15-$\lambda$ factor out of the determinant. Therefore one real eigen value would be 15.

Comments

why did you add columns like this? which property of determinant is this?

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Property #14 from wikipedia. https://en.wikipedia.org/wiki/Determinant

Adding a scalar multiple of one column to another column does not change the value of the determinant. This is a consequence of properties 8 and 10 in the following way: by property 8 the determinant changes by a multiple of the determinant of a matrix with two equal columns, which determinant is 0 by property 10. Similarly, adding a scalar multiple of one row to another row leaves the determinant unchanged.

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On applying rows/ column operation on the matrix |$ A- \lambda I$| do we have same set of eigenvalues as before?