1
Eigenvalues and Eigenvectors
Calculations involving matrices can become quite messy. For example, if we need
to calculate powers of an 𝑛 × 𝑛 matrix 𝐴, this can be cumbersome. However, if
we can find a basis for which 𝐴 is a diagonal matrix (ie 𝐴𝑖𝑗 = 0 if 𝑖 ≠ 𝑗 ) then
caclulations become easier.
Def. A linear operator 𝑇 (ie a linear transformation mapping 𝑉 → 𝑉) on a finite
dimensional vector space 𝑉 is called diagonalizable if there is an ordered basis 𝐵
of 𝑉 such that [𝑇]𝐵 is a diagonal matrix. A square matrix is called diagonalizable if
𝐿𝐴 is diagonalizable.
Notice that if 𝐵 = {𝑣1 , … , 𝑣𝑛 } is an ordered basis for 𝑉 for which 𝑇: 𝑉 → 𝑉 is
diagonalizable then if 𝐴 = [𝑇]𝐵 and 𝑣𝑗 ∈ 𝐵 we have
𝑇(𝑣𝑗 ) = ∑𝑛𝑖=1 𝐴𝑖𝑗 𝑣𝑖 = 𝐴𝑗𝑗 𝑣𝑗 = 𝜆𝑗 𝑣𝑗 , where 𝜆𝑗 = 𝐴𝑗𝑗 .
Conversely, if 𝐵 = {𝑣1 , … , 𝑣𝑛 } is an ordered basis for 𝑉 such that
𝑇(𝑣𝑗 ) = 𝜆𝑗 𝑣𝑗 , for 𝜆𝑗 ∈ ℝ then
𝜆1 0 0 ⋯ 0
0 𝜆2 0 ⋯ 0
[𝑇]𝐵 = 0 0 𝜆3 ⋯ 0 .
0 0 0 ⋱ ⋮
[0 0 0 ⋯ 𝜆𝑛 ]
Def. Let 𝑇 be a linear operator on a vector space 𝑉. A nonzero vector 𝑣 ∈ 𝑉 is
called an eigenvector of 𝑇 if there exists a 𝜆 ∈ ℝ such that 𝑇(𝑣) = 𝜆𝑣. 𝜆 is called
the eigenvalue of 𝑇 corresponding to 𝑣.
, 2
So a linear operator 𝑇: 𝑉 → 𝑉, 𝑉 a finite dimensional vector space, is
diagonalizable if and only if there exists an ordered basis 𝐵 = {𝑣1 , … , 𝑣𝑛 } for 𝑉 of
eigenvectors of 𝑇.
1 3 1 3
Ex. Let 𝐴 = [ ], 𝑣1 = [ ], and 𝑣2 = [ ]. Show that 𝑣1 and 𝑣2 are
4 2 −1 4
eigenvectors of 𝐴.
1 3 1 −2 1
𝐴𝑣1 = [ ][ ] = [ ] = −2 [ ] = −2𝑣1
4 2 −1 2 −1
1 3 3 15 3
𝐴𝑣2 = [ ] [ ] = [ ] = 5 [ ] = 5𝑣2 .
4 2 4 20 4
So −2 is the eigenvalue corresponding to the eigenvector 𝑣1
and 5 is the eigenvalue corresponding to the eigenvector 𝑣2 .
Thus with respect to the basis 𝐵′ = {< 1, −1 >, < 3, 4 >} we have
−2 0
𝐴=[ ].
0 5
Notice that if we had used the change of basis formula, 𝑃−1 𝐴𝑃, for changing the
basis for 𝐴 from the standard basis {< 1,0 >, < 0,1 >} to
𝐵′ = {< 1, −1 >, < 3, 4 >} we would get
1 3 1 4 −3 1 4 −3
𝑃=[ ] ⟹ 𝑃−1 = [ ]= [ ], and
−1 4 det(𝑃) 1 1 7 1 1
14 −3 1 3 1 3
𝑃−1 𝐴𝑃 = [ ][ ][ ]
7 1 1 4 2 −1 4
1 4 −3 −2 15 1 −14 0 −2 0
= [ ][ ]= [ ]=[ ].
7 1 1 2 20 7 0 35 0 5
Eigenvalues and Eigenvectors
Calculations involving matrices can become quite messy. For example, if we need
to calculate powers of an 𝑛 × 𝑛 matrix 𝐴, this can be cumbersome. However, if
we can find a basis for which 𝐴 is a diagonal matrix (ie 𝐴𝑖𝑗 = 0 if 𝑖 ≠ 𝑗 ) then
caclulations become easier.
Def. A linear operator 𝑇 (ie a linear transformation mapping 𝑉 → 𝑉) on a finite
dimensional vector space 𝑉 is called diagonalizable if there is an ordered basis 𝐵
of 𝑉 such that [𝑇]𝐵 is a diagonal matrix. A square matrix is called diagonalizable if
𝐿𝐴 is diagonalizable.
Notice that if 𝐵 = {𝑣1 , … , 𝑣𝑛 } is an ordered basis for 𝑉 for which 𝑇: 𝑉 → 𝑉 is
diagonalizable then if 𝐴 = [𝑇]𝐵 and 𝑣𝑗 ∈ 𝐵 we have
𝑇(𝑣𝑗 ) = ∑𝑛𝑖=1 𝐴𝑖𝑗 𝑣𝑖 = 𝐴𝑗𝑗 𝑣𝑗 = 𝜆𝑗 𝑣𝑗 , where 𝜆𝑗 = 𝐴𝑗𝑗 .
Conversely, if 𝐵 = {𝑣1 , … , 𝑣𝑛 } is an ordered basis for 𝑉 such that
𝑇(𝑣𝑗 ) = 𝜆𝑗 𝑣𝑗 , for 𝜆𝑗 ∈ ℝ then
𝜆1 0 0 ⋯ 0
0 𝜆2 0 ⋯ 0
[𝑇]𝐵 = 0 0 𝜆3 ⋯ 0 .
0 0 0 ⋱ ⋮
[0 0 0 ⋯ 𝜆𝑛 ]
Def. Let 𝑇 be a linear operator on a vector space 𝑉. A nonzero vector 𝑣 ∈ 𝑉 is
called an eigenvector of 𝑇 if there exists a 𝜆 ∈ ℝ such that 𝑇(𝑣) = 𝜆𝑣. 𝜆 is called
the eigenvalue of 𝑇 corresponding to 𝑣.
, 2
So a linear operator 𝑇: 𝑉 → 𝑉, 𝑉 a finite dimensional vector space, is
diagonalizable if and only if there exists an ordered basis 𝐵 = {𝑣1 , … , 𝑣𝑛 } for 𝑉 of
eigenvectors of 𝑇.
1 3 1 3
Ex. Let 𝐴 = [ ], 𝑣1 = [ ], and 𝑣2 = [ ]. Show that 𝑣1 and 𝑣2 are
4 2 −1 4
eigenvectors of 𝐴.
1 3 1 −2 1
𝐴𝑣1 = [ ][ ] = [ ] = −2 [ ] = −2𝑣1
4 2 −1 2 −1
1 3 3 15 3
𝐴𝑣2 = [ ] [ ] = [ ] = 5 [ ] = 5𝑣2 .
4 2 4 20 4
So −2 is the eigenvalue corresponding to the eigenvector 𝑣1
and 5 is the eigenvalue corresponding to the eigenvector 𝑣2 .
Thus with respect to the basis 𝐵′ = {< 1, −1 >, < 3, 4 >} we have
−2 0
𝐴=[ ].
0 5
Notice that if we had used the change of basis formula, 𝑃−1 𝐴𝑃, for changing the
basis for 𝐴 from the standard basis {< 1,0 >, < 0,1 >} to
𝐵′ = {< 1, −1 >, < 3, 4 >} we would get
1 3 1 4 −3 1 4 −3
𝑃=[ ] ⟹ 𝑃−1 = [ ]= [ ], and
−1 4 det(𝑃) 1 1 7 1 1
14 −3 1 3 1 3
𝑃−1 𝐴𝑃 = [ ][ ][ ]
7 1 1 4 2 −1 4
1 4 −3 −2 15 1 −14 0 −2 0
= [ ][ ]= [ ]=[ ].
7 1 1 2 20 7 0 35 0 5
