1
Diagonalizability
So far we know that a linear operator 𝑇 on 𝑉 or its associated matrix is
diagonalizable if and only if there exists an ordered basis 𝐵 = {𝑣1 , … , 𝑣𝑛 } of
eigenvectors of 𝑇. However, we don’t know yet when a basis of eigenvectors
1 1
exists. We saw in an example (𝐴 = [ ] ) that there are linear
0 1
operators/matrices which are not diagonalizable.
Theorem: Let 𝑇 be a linear operator on a vector space 𝑉, and let 𝜆1 , … , 𝜆𝑘 be
distinct eigenvalues of 𝑇. If 𝑣1 , … , 𝑣𝑘 are eigenvectors of 𝑇 such that 𝜆𝑖
corresponds to 𝑣𝑖 for 1 ≤ 𝑖 ≤ 𝑘, then {𝑣1 , … , 𝑣𝑘 } is linearly independent.
Proof: The proof is by induction on 𝑘.
For 𝑘 = 1, 𝑣1 ≠ 0 since it’s an eigenvector so {𝑣1 } is linearly independent.
Now assume the theorem holds for 𝑘 − 1 distinct eigenvalues and let’s
prove it for 𝑘 distinct eigenvalues.
Let {𝑣1 , … 𝑣𝑘 } be eigenvectors associated with the distinct eigenvalues
𝜆1 , … , 𝜆𝑘 .
Suppose {𝑣1 , … , 𝑣𝑘 } is linearly dependent. Then we have:
(∗) 𝑎1 𝑣1 + ⋯ 𝑎𝑘 𝑣𝑘 = 0; where not all of the 𝑎𝑖 ′𝑠 are 0.
Let’s apply 𝑇 − 𝜆𝑘 𝐼 to both sides of this equations:
(𝑇 − 𝜆𝑘 𝐼)(𝑎1 𝑣1 + ⋯ 𝑎𝑘 𝑣𝑘 ) = 0
𝑎1 (𝑇 − 𝜆𝑘 𝐼)𝑣1 + ⋯ 𝑎𝑘 (𝑇 − 𝜆𝑘 𝐼)𝑣𝑘 = 0
𝑎1 (𝜆1 − 𝜆𝑘 )𝑣1 + ⋯ + 𝑎𝑘−1 (𝜆𝑘−1 − 𝜆𝑘 )𝑣𝑘−1 + 𝑎𝑘 (𝜆𝑘 − 𝜆𝑘 ) = 0.
So 𝑎1 (𝜆1 − 𝜆𝑘 )𝑣1 + ⋯ + 𝑎𝑘−1 (𝜆𝑘−1 − 𝜆𝑘 )𝑣𝑘−1 = 0.
, 2
But {𝑣1 , … , 𝑣𝑘−1 } is linearly independent so 𝑎1 , … , 𝑎𝑘−1 = 0, since 𝜆𝑖 ≠ 𝜆𝑘 if
𝑖 ≠ 𝑘.
Thus from (∗) we get 𝑎𝑘 = 0.
Hence {𝑣1 , … , 𝑣𝑘 } is linearly independent.
Corollary: Let 𝑇 be a linear operator on an 𝑛-dimensional vector space 𝑉. If 𝑇 has
𝑛 distinct eigenvalues then 𝑇 is diagonalizable.
The converse to the previous theorem is false. That is, if 𝑇 is diagonalizable it is
not true that 𝑇 must have 𝑛 distinct eigenvalues. For example, the identity linear
operator is diagonal but has only one distinct eigenvalue.
If 𝑇 is a linear operator on an 𝑛-dimensional vector space 𝑉, then its characteristic
polynomial, 𝑝(𝜆) = det(𝐴 − 𝜆𝐼), where [𝑇] = 𝐴, is a polynomial of degree 𝑛 in 𝜆.
In general not every polynomial of degree 𝑛 can be completely factored into
linear factors with real coefficients. For example, 𝑝(𝜆) = 𝜆2 + 1 can’t be factored
into linear factors with real coefficients (it can with complex coefficients). If a
polynomial of degree 𝑛 factors completely into linear factors with real
coefficients, ie
𝑝(𝜆) = 𝑐(𝜆 − 𝑎1 ) ⋯ (𝜆 − 𝑎𝑛 )
then we say 𝒑(𝝀) splits over ℝ. Note that the 𝑎𝑖 ’s need not be distinct.
Diagonalizability
So far we know that a linear operator 𝑇 on 𝑉 or its associated matrix is
diagonalizable if and only if there exists an ordered basis 𝐵 = {𝑣1 , … , 𝑣𝑛 } of
eigenvectors of 𝑇. However, we don’t know yet when a basis of eigenvectors
1 1
exists. We saw in an example (𝐴 = [ ] ) that there are linear
0 1
operators/matrices which are not diagonalizable.
Theorem: Let 𝑇 be a linear operator on a vector space 𝑉, and let 𝜆1 , … , 𝜆𝑘 be
distinct eigenvalues of 𝑇. If 𝑣1 , … , 𝑣𝑘 are eigenvectors of 𝑇 such that 𝜆𝑖
corresponds to 𝑣𝑖 for 1 ≤ 𝑖 ≤ 𝑘, then {𝑣1 , … , 𝑣𝑘 } is linearly independent.
Proof: The proof is by induction on 𝑘.
For 𝑘 = 1, 𝑣1 ≠ 0 since it’s an eigenvector so {𝑣1 } is linearly independent.
Now assume the theorem holds for 𝑘 − 1 distinct eigenvalues and let’s
prove it for 𝑘 distinct eigenvalues.
Let {𝑣1 , … 𝑣𝑘 } be eigenvectors associated with the distinct eigenvalues
𝜆1 , … , 𝜆𝑘 .
Suppose {𝑣1 , … , 𝑣𝑘 } is linearly dependent. Then we have:
(∗) 𝑎1 𝑣1 + ⋯ 𝑎𝑘 𝑣𝑘 = 0; where not all of the 𝑎𝑖 ′𝑠 are 0.
Let’s apply 𝑇 − 𝜆𝑘 𝐼 to both sides of this equations:
(𝑇 − 𝜆𝑘 𝐼)(𝑎1 𝑣1 + ⋯ 𝑎𝑘 𝑣𝑘 ) = 0
𝑎1 (𝑇 − 𝜆𝑘 𝐼)𝑣1 + ⋯ 𝑎𝑘 (𝑇 − 𝜆𝑘 𝐼)𝑣𝑘 = 0
𝑎1 (𝜆1 − 𝜆𝑘 )𝑣1 + ⋯ + 𝑎𝑘−1 (𝜆𝑘−1 − 𝜆𝑘 )𝑣𝑘−1 + 𝑎𝑘 (𝜆𝑘 − 𝜆𝑘 ) = 0.
So 𝑎1 (𝜆1 − 𝜆𝑘 )𝑣1 + ⋯ + 𝑎𝑘−1 (𝜆𝑘−1 − 𝜆𝑘 )𝑣𝑘−1 = 0.
, 2
But {𝑣1 , … , 𝑣𝑘−1 } is linearly independent so 𝑎1 , … , 𝑎𝑘−1 = 0, since 𝜆𝑖 ≠ 𝜆𝑘 if
𝑖 ≠ 𝑘.
Thus from (∗) we get 𝑎𝑘 = 0.
Hence {𝑣1 , … , 𝑣𝑘 } is linearly independent.
Corollary: Let 𝑇 be a linear operator on an 𝑛-dimensional vector space 𝑉. If 𝑇 has
𝑛 distinct eigenvalues then 𝑇 is diagonalizable.
The converse to the previous theorem is false. That is, if 𝑇 is diagonalizable it is
not true that 𝑇 must have 𝑛 distinct eigenvalues. For example, the identity linear
operator is diagonal but has only one distinct eigenvalue.
If 𝑇 is a linear operator on an 𝑛-dimensional vector space 𝑉, then its characteristic
polynomial, 𝑝(𝜆) = det(𝐴 − 𝜆𝐼), where [𝑇] = 𝐴, is a polynomial of degree 𝑛 in 𝜆.
In general not every polynomial of degree 𝑛 can be completely factored into
linear factors with real coefficients. For example, 𝑝(𝜆) = 𝜆2 + 1 can’t be factored
into linear factors with real coefficients (it can with complex coefficients). If a
polynomial of degree 𝑛 factors completely into linear factors with real
coefficients, ie
𝑝(𝜆) = 𝑐(𝜆 − 𝑎1 ) ⋯ (𝜆 − 𝑎𝑛 )
then we say 𝒑(𝝀) splits over ℝ. Note that the 𝑎𝑖 ’s need not be distinct.
