The eigenvalues and eigenvectors have been defined in terms of linear transformations and vectors. In this section, we give matrix interpretation.
Let A be an n by n matrix. Let v_{1}, v_{2}, ..., v_{k} be vectors and let λ_{1}, λ_{2}, ..., λ_{k} be numbers. Denote
P = [v_{1} v_{2} ... v_{k}], D = [  λ_{1}  0  . .  0  ]. 
0  λ_{2}  . .  0  
:  :  :  
0  0  . .  λ_{k} 
Then the equalities
Av_{1} = λ_{1}v_{1}, Av_{2} = λ_{2}v_{2}, ..., Av_{k} = λ_{k}v_{k},
AP = [Av_{1} Av_{2} ... Av_{k}] = [λ_{1}v_{1} λ_{2}v_{2} ... λ_{k}v_{k}] = [v_{1} v_{2} ... v_{k}]D = PD.
Moreover, v_{1}, v_{2}, ..., v_{k} form a basis if and only if P is an invertible matrix, and the equality AP = PD becomes A = PDP^{1}.
We just proved that the two properties in the following definition are equivalent.
A matrix is diagonalizable if it has any of the following equivalent properties
Moreover, any expression A = PDP^{1} is a diagonalization of A.
The first property can also be used to define the diagonalizability for linear transformations T: V → V. Then the second property has the following interpretation. Let A = [T(B)]_{B} be the matrix of T with respect to a basis B = {b_{1}, b_{2}, ..., b_{n}} of V. Let E = {v_{1}, v_{2}, ..., v_{n}} be a basis of eigenvectors for T. The fact T(v_{i}) = λ_{i}v_{i} simply means that the matrix D = [T(E)]_{E} of T with respect to the basis E is diagonal. By this discussion, we conclude
A = [T(B)]_{B} = [E]_{B} [T(E)]_{E} [E]_{B}^{1} = PDP^{1}.
Example In an earlier example, we found the matrix
A = [  13  4  ] 
4  7 
has a basis v_{1} = (1, 2), v_{2} = (2, 1) of eigenvectors, with eigenvalues λ_{1} = 5, λ_{2} = 15. Thus A = PDP^{1}, with
P = [v_{1} v_{2}] = [  1  2  ], D = [  λ_{1}  0  ] = [  5  0  ]. 
2  1  0  λ_{2}  0  15 
In another earlier example, we found the matrix
[  1  3  3  ] 
3  7  3  
6  6  2 
has a basis (1, 1, 0), (1, 0, 1), (1, 1, 2) of eigenvectors, with corresponding eigenvalues 4, 4, 2. This gives us
[  1  3  3  ] = [  1  1  1  ] [  4  0  0  ] [  1  1  1  ]^{1}. 
3  7  3  1  0  1  0  4  0  1  0  1  
6  6  2  0  1  2  0  0  2  0  1  2 
Note that by rearranging the order of the eigenvectors in the basis, with corresponding change of order for eigenvalues, we have other diagonalizations of the same matrix
[  1  3  3  ] = [  1  1  1  ] [  2  0  0  ] [  1  1  1  ]^{1}, 
3  7  3  1  0  1  0  4  0  1  0  1  
6  6  2  2  1  0  0  0  4  2  1  0 
[  1  3  3  ] = [  1  1  1  ] [  4  0  0  ] [  1  1  1  ]^{1}. 
3  7  3  1  1  0  0  2  0  1  1  0  
6  6  2  0  2  1  0  0  4  0  2  1 
A = [  2  3  3  ]. 
0  2  3  
0  0  1 
Being upper triangular, it is easy to see that the characteristic polynomial is (2  λ)^{2}(1  λ) and there are two eigenvalues λ_{1} = 2, λ_{2} = 1.
For λ_{1} = 2, we find v_{1} = (1, 0, 0) form a basis of nul(A  2I). For λ_{2} = 1, we find v_{2} = (6, 3, 1) form a basis of nul(A  I). Since we only get two eigenvectors, A does not have a basis of eigenvectors and is not diagonalizable.
In terms of matrices, we have
AP = [  2  3  3  ] [  1  6  ] =  [  1  6  ] 


0  2  3  0  3  0  3  
0  0  1  0  1  0  1 
However, the matrix
P = [  1  6  ] 
0  3  
0  1 
is not invertible, and we cannot multiply P^{1} on the right of AP = PD to get diagonalization.