# Properties of Eigenvalues

Andrew Thangaraj

Aug-Nov 2020

## Recap

- Vector space \(V\) over a scalar field \(F\)
- \(F\): real field \(\mathbb{R}\) or complex field \(\mathbb{C}\) in this course

- \(m\times n\) matrix A represents a linear map \(T:F^n\to F^m\)
- dim null \(T\) + dim range \(T\) = dim \(V\)

- Linear equation: \(Ax=b\)
- Solution (if it exists): \(u+\) null\((A)\)

- Four fundamental subspaces of a matrix
- Column space, row space, null space, left null space

- Eigenvalue \(\lambda\) and Eigenvector \(v\): \(Tv=\lambda v\)
- Distinct eigenvalues have independent eigenvectors
- Basis of eigenvectors results in a diagonal matrix for \(T\)
- Every linear map has an upper triangular matrix representation

## Eigenvalues, nullity and invertibility

\(T:V\to V\)

Zero is an eigenvalue iff dim null \(T\) > 0.

\(v\): eigenvector with eigenvalue 0 \(\Leftrightarrow\) \(Tv=0\), \(v\ne0\)

\(T\) is invertible iff no eigenvalue is zero.

Corollary of above

\(T\): invertible with eigenvalue \(\lambda\ne0\), eigenvector \(v\). Then, \(1/\lambda\) is an eigenvalue of \(T^{-1}\) with eigenvector \(v\).

\(Tv=\lambda v\) implies \(T^{-1}v=(1/\lambda)v\)

## Eigenvalues and transpose

Suppose \(\lambda\) is an eigenvalue of \(A\). Then, \(\lambda\) is an eigenvalue of \(A^T\).

*Proof*

\(\lambda\): \(A-\lambda I\) is non-invertible

rank\((A-\lambda I)=\) rank\((A^T-\lambda I)\)

So, \(A^T-\lambda I\) is non-invertible

## Eigenvalues and determinant

Determinant is equal to the product of eigenvalues.

*Proof*

Find upper triangular matrix representation for \(T\).

Determinant is product of diagonal entries, which are eigenvalues.

## Trace of an operator

\(T:V\to V\), \(A\): matrix w.r.t. some basis

Trace of \(T\), denoted tr\((T)\), is defined as sum of diagonal elements of \(A\)

*why is definition valid?*

Sum of diagonal elements of \(SAS^{-1}\) and \(A\) are equal.

*Proof*

## Eigenvalues and trace

Trace is equal to the sum of eigenvalues.

*Proof*

Find upper triangular matrix representation for \(T\).

Trace is sum of diagonal entries, which are eigenvalues.