Andrew Thangaraj
Aug-Nov 2020
\(T:V\to V\)
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 \(T^{-1}v=(1/\lambda)v\)
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
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.
\(T:V\to V\), \(A\): 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
tr\((AB) =\) tr\((BA)\) (exercise)
tr\((SAS^{-1})=\) tr\((S(AS^{-1}))=\) tr\(((AS^{-1})S)=\) tr\((A)\)
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.