Simplifying matrices of linear maps by choice of basis

Andrew Thangaraj

Aug-Nov 2020


  • 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
  • Determinant of a square matrix
    • Function with many interesting properties
  • Change of basis for linear map
    • (change-basis in \(W\)) (linear map) (change-basis in \(V\))

“Simple” matrix

\(n\times n\): square matrix

  • Simplest is identity

  • Diagonal matrices

  • As many zeroes as possible off diagonal

\(m\times n\) matrices

  • Non-zero entries only in main diagonal

“Simple” matrices for linear maps

\(T:V\to W\), dim \(V=n\), dim \(W=m\)

Basis for \(V\)

  • Basis of null \(T\): \(\{u_1,\ldots,u_{n-r}\}\), extend to basis of \(V\)

  • \(B_V=\{v_1,\ldots,v_r,u_1,\ldots,u_{n-r}\}\), \(r\): rank

Basis for \(W\)

  • Basis of range \(T\): \(\{T(v_1),\ldots,T(v_r)\}\), extend to basis of \(W\)

  • \(B_W=\{T(v_1),\ldots,T(v_r),w_1,\ldots,w_{m-r}\}\)

Matrix of \(T\) w.r.t. \(B_V\), \(B_W\)

\(\begin{bmatrix} \vdots&\cdots&\vdots&\vdots&\cdots&\vdots\\ T(v_1)&\cdots&T(v_r)&T(u_1)&\cdots&T(u_{n-r})\\ \vdots&\cdots&\vdots&\vdots&\cdots&\vdots \end{bmatrix}=\begin{bmatrix} I_r&0\\ 0&0 \end{bmatrix}\)

Elementary row-column operations and change of basis

\(A:m\times n\) matrix, rank: \(r\), represents \(T:F^n\to F^m\) in standard basis

There exist elementary row operations \(E_i\), col operations \(F_j\) s.t.

\(\left(\prod_i E_i\right) A \left(\prod_jF_j\right)=\begin{bmatrix} I_r&0\\ 0&0 \end{bmatrix}\)

\(S_L=\prod_i E_i\): \(m\times m\) invertible

\(S_R=\prod_jF_j\): \(n\times n\) invertible

\(A=S^{-1}_L\begin{bmatrix} I_r&0\\ 0&0 \end{bmatrix}S^{-1}_R\)

Change of basis

  • Basis for \(F^n\): columns of \(S_R\)

  • Basis for \(F^m\): columns of \(S^{-1}_L\)

\(T\leftrightarrow\begin{bmatrix} I_r&0\\ 0&0 \end{bmatrix}\)

What about operators?

\(A:n\times n\) matrix, rank: \(r\), represents \(T:F^n\to F^n\) in standard basis

Important constraint: same basis for input and Output

Goal: Find invertible \(S\) s.t. \(S^{-1}AS\) is “simple”

  • Same vector space, same basis

  • Operators are typically used multiple times on a vector

\(Av\), \(A^2v\), \(A^3v\),\(\ldots\)

  • Change of basis with same basis for input and output

\(S^{-1}ASv\), \((S^{-1}AS)^2v=S^{-1}A^2Sv\),\(\ldots\)

  • Well-behaved when operator is repeatedly applied

Rest of course: study operators in above scenario