Effect of Basis on Linear Transformation Matrix

Algebra Level 2

Let T : V → V T: V \to V be a linear transformation in a finite-dimensional of a vector space . Then, for some choice of basis B \mathcal{B} and matrix M B M_\mathcal{B} , T ( v ) = M B v T(v) = M_\mathcal{B}v . Which properties of the matrix M B M_\mathcal{B} remain unchanged regardless of basis B ? \mathcal{B}?

Rank Invertibility, rank Invertibility, value of determinant Value of determinant, rank Invertibility, value of determinant, rank

Henry Maltby by Henry Maltby , 26, USA

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1 solution

Henry Maltby
Jul 7, 2016

Only the invertibility and rank are independent of the choice of matrix.

A linear transformation is invertible if and only if its matrix is. So invertibility is independent of the choice of basis. Similarly, the size of the space into which the linear transformation is fixed, independent of basis. So rank is independent of the choice of basis.

The value of the determinant can be changed by simply taking a multiple of a basis element. The only information that can be gleamed from the determinant is whether it is nonzero (i.e., invertible). The value of the determinant is not independent of basis.

3 Helpful 1 Interesting 0 Brilliant 0 Confused

The determinant is independent of the choice of basis. One way to understand why is because it's the product of eigenvalues which are independent of the choice of basis. Another one is: let A be the matrix representation of a linear transformation, this matrix representation will be similar to the same matrix representation on any other basis, that is : there is some invertible matrix Q (a change of basis matrix) such that : A=QBQ^(-1) where B is the matrix representation of the same linear transformation on that different basis. It can be easily shown that the determinants of A and B are exactly the same.

Alexandros Kazantzidis - 4 years, 10 months ago

sorry i don't understand why the determinant of the matrix remains the same? eg. 2x2 diagnol matrices all satisfy R2 ->R2 but their determinants aren't the same. Thanks for anyone answer!

呈翔 马 - 2 years ago

Explanation is correct, but the option given in answer is incorrect. Answer options should be 'invertibility and rank'

surya prakash Sahu - 12 months ago

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yes, i think ur right. can this be changed?

Dhruv G - 4 months, 2 weeks ago

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