Equivalence and similarity of matrices

Sarang S. Sane
Equivalence of matrices
Let A and B be two matrices of order m × n. We say A is equivalent
to B if B = QAP for some invertible n × n matrix P and for some
invertible m × m matrix Q.
Equivalence of matrices
Let A and B be two matrices of order m × n. We say A is equivalent
to B if B = QAP for some invertible n × n matrix P and for some
invertible m × m matrix Q.

Other characterisations:
1) A can be transformed into B by a combination of elementary
row and column operations.
Equivalence of matrices
Let A and B be two matrices of order m × n. We say A is equivalent
to B if B = QAP for some invertible n × n matrix P and for some
invertible m × m matrix Q.

Other characterisations:
1) A can be transformed into B by a combination of elementary
row and column operations.

2) rank(A) = rank(B)
Equivalence of matrices
Let A and B be two matrices of order m × n. We say A is equivalent
to B if B = QAP for some invertible n × n matrix P and for some
invertible m × m matrix Q.

Other characterisations:
1) A can be transformed into B by a combination of elementary
row and column operations.

2) rank(A) = rank(B)

Equivalence of matrices is an equivalence relation

Equivalence of matrices
Let A and B be two matrices of order m × n. We say A is equivalent
to B if B = QAP for some invertible n × n matrix P and for some
invertible m × m matrix Q.

Other characterisations:
1) A can be transformed into B by a combination of elementary
row and column operations.

2) rank(A) = rank(B)

Equivalence of matrices is an equivalence relation i.e.

I A is equivalent to itself
Equivalence of matrices
Let A and B be two matrices of order m × n. We say A is equivalent
to B if B = QAP for some invertible n × n matrix P and for some
invertible m × m matrix Q.

Other characterisations:
1) A can be transformed into B by a combination of elementary
row and column operations.

2) rank(A) = rank(B)

Equivalence of matrices is an equivalence relation i.e.

I A is equivalent to itself
I A is equivalent to B implies B is equivalent to A.
Equivalence of matrices
Let A and B be two matrices of order m × n. We say A is equivalent
to B if B = QAP for some invertible n × n matrix P and for some
invertible m × m matrix Q.

Other characterisations:
1) A can be transformed into B by a combination of elementary
row and column operations.

2) rank(A) = rank(B)

Equivalence of matrices is an equivalence relation i.e.

I A is equivalent to itself
I A is equivalent to B implies B is equivalent to A.
I A is equivalent to B and B to C implies A is equivalent to C .
Equivalence of matrices
Let A and B be two matrices of order m × n. We say A is equivalent
to B if B = QAP for some invertible n × n matrix P and for some
invertible m × m matrix Q.

Other characterisations:
1) A can be transformed into B by a combination of elementary
row and column operations.

2) rank(A) = rank(B)

Equivalence of matrices is an equivalence relation i.e.

I A is equivalent to itself
I A is equivalent to B implies B is equivalent to A.
I A is equivalent to B and B to C implies A is equivalent to C .
Example
Consider the linear transformation f : R3 → R2 , defined as :
f (x, y , z) = (x + y , y + z).
Example
Consider the linear transformation f : R3 → R2 , defined as :
f (x, y , z) = (x + y , y + z).

Consider two ordered bases for R3 :

β1 = (1, 0, 0), (0, 1, 0), (0, 0, 1) and β2 = (1, 1, 0), (0, 1, 1), (0, 0, 1).
Example
Consider the linear transformation f : R3 → R2 , defined as :
f (x, y , z) = (x + y , y + z).

Consider two ordered bases for R3 :

β1 = (1, 0, 0), (0, 1, 0), (0, 0, 1) and β2 = (1, 1, 0), (0, 1, 1), (0, 0, 1).

Similarly, consider two ordered bases for R2 :

γ1 = (1, 0), (0, 1) and γ2 = (1, 0), (1, 1).
Example
Consider the linear transformation f : R3 → R2 , defined as :
f (x, y , z) = (x + y , y + z).

Consider two ordered bases for R3 :

β1 = (1, 0, 0), (0, 1, 0), (0, 0, 1) and β2 = (1, 1, 0), (0, 1, 1), (0, 0, 1).

Similarly, consider two ordered bases for R2 :

γ1 = (1, 0), (0, 1) and γ2 = (1, 0), (1, 1).

f (1, 0, 0) = (1, 0),

f (0, 1, 0) = (1, 1) = 1(1, 0) + 1(0, 1),
f (0, 0, 1) = (0, 1).
Example
Consider the linear transformation f : R3 → R2 , defined as :
f (x, y , z) = (x + y , y + z).

Consider two ordered bases for R3 :

β1 = (1, 0, 0), (0, 1, 0), (0, 0, 1) and β2 = (1, 1, 0), (0, 1, 1), (0, 0, 1).

Similarly, consider two ordered bases for R2 :

γ1 = (1, 0), (0, 1) and γ2 = (1, 0), (1, 1).

f (1, 0, 0) = (1, 0),

f (0, 1, 0) = (1, 1) = 1(1, 0) + 1(0, 1),
f (0, 0, 1) = (0, 1).

Hence the matrix

 corresponding
 to f with respect to the bases β1
1 1 0
and γ1 is A =
0 1 1
Example (contd.)
f (1, 1, 0) = (2, 1) = 1(1, 0) + 1(1, 1),
f (0, 1, 1) = (1, 2) = −1(1, 0) + 2(1, 1),
f (0, 0, 1) = (0, 1) = −1(1, 0) + 1(1, 1).
Example (contd.)
f (1, 1, 0) = (2, 1) = 1(1, 0) + 1(1, 1),
f (0, 1, 1) = (1, 2) = −1(1, 0) + 2(1, 1),
f (0, 0, 1) = (0, 1) = −1(1, 0) + 1(1, 1).

Hence the matrix

 corresponding
 to f with respect to the bases β2
1 −1 −1
and γ2 is B = .
1 2 1
Example (contd.)
f (1, 1, 0) = (2, 1) = 1(1, 0) + 1(1, 1),
f (0, 1, 1) = (1, 2) = −1(1, 0) + 2(1, 1),
f (0, 0, 1) = (0, 1) = −1(1, 0) + 1(1, 1).

Hence the matrix

 corresponding
 to f with respect to the bases β2
1 −1 −1
and γ2 is B = .
1 2 1
 
1 0 0  
1 −1
Choose P = 1 1 0 and Q =
  .
0 1
0 1 1
Example (contd.)
f (1, 1, 0) = (2, 1) = 1(1, 0) + 1(1, 1),
f (0, 1, 1) = (1, 2) = −1(1, 0) + 2(1, 1),
f (0, 0, 1) = (0, 1) = −1(1, 0) + 1(1, 1).

Hence the matrix

 corresponding
 to f with respect to the bases β2
1 −1 −1
and γ2 is B = .
1 2 1
 
1 0 0  
1 −1
Choose P = 1 1 0 and Q =
  . Then
0 1
0 1 1
 
   1 0 0  
1 −1 1 1 0  1 −1 −1
QAP = 1 1 0 = =B
0 1 0 1 1 1 2 1
0 1 1
.
Hence A and B are equivalent to each other.
Linear transformations and equivalence of matrices

Consider a linear transformation T : V → W ,

Linear transformations and equivalence of matrices

Consider a linear transformation T : V → W , two ordered bases

β1 and β2 for V ,
Linear transformations and equivalence of matrices

Consider a linear transformation T : V → W , two ordered bases

β1 and β2 for V , and two ordered bases γ1 and γ2 for W .
Linear transformations and equivalence of matrices

Consider a linear transformation T : V → W , two ordered bases

β1 and β2 for V , and two ordered bases γ1 and γ2 for W .

Let A be the matrix corresponding to T with respect to the bases

β1 and γ1
Linear transformations and equivalence of matrices

Consider a linear transformation T : V → W , two ordered bases

β1 and β2 for V , and two ordered bases γ1 and γ2 for W .

Let A be the matrix corresponding to T with respect to the bases

β1 and γ1 and B be the matrix corresponding to T with respect to
the bases β2 and γ2 .
Linear transformations and equivalence of matrices

Consider a linear transformation T : V → W , two ordered bases

β1 and β2 for V , and two ordered bases γ1 and γ2 for W .

Let A be the matrix corresponding to T with respect to the bases

β1 and γ1 and B be the matrix corresponding to T with respect to
the bases β2 and γ2 .

Then A is equivalent to B !
Similar matrices

An n × n matrix A is similar to an n × n matrix B if there exists an

n × n invertible matrix P such that B = P −1 AP.
Similar matrices

An n × n matrix A is similar to an n × n matrix B if there exists an

n × n invertible matrix P such that B = P −1 AP.

Note that similarity is an equivalence relation,

Similar matrices

An n × n matrix A is similar to an n × n matrix B if there exists an

n × n invertible matrix P such that B = P −1 AP.

Note that similarity is an equivalence relation, i.e. :

I A is similar to itself
Similar matrices

An n × n matrix A is similar to an n × n matrix B if there exists an

n × n invertible matrix P such that B = P −1 AP.

Note that similarity is an equivalence relation, i.e. :

I A is similar to itself
I A is similar to B implies B is similar to A.
Similar matrices

An n × n matrix A is similar to an n × n matrix B if there exists an

n × n invertible matrix P such that B = P −1 AP.

Note that similarity is an equivalence relation, i.e. :

I A is similar to itself
I A is similar to B implies B is similar to A.
I A is similar to B and B to C implies A is similar to C .
Similar matrices

An n × n matrix A is similar to an n × n matrix B if there exists an

n × n invertible matrix P such that B = P −1 AP.

Note that similarity is an equivalence relation, i.e. :

I A is similar to itself
I A is similar to B implies B is similar to A.
I A is similar to B and B to C implies A is similar to C .
Important properties of similar matrices

Suppose A and B are similar matrices. Then the following

properties hold :
Important properties of similar matrices

Suppose A and B are similar matrices. Then the following

properties hold :

I A and B are equivalent.

Important properties of similar matrices

Suppose A and B are similar matrices. Then the following

properties hold :

I A and B are equivalent.

I A and B have the same rank.

Important properties of similar matrices

Suppose A and B are similar matrices. Then the following

properties hold :

I A and B are equivalent.

I A and B have the same rank.

I det(B) = det(P −1 AP) = det(P −1 )det(A)det(P)

1
= det(A)det(P) = det(A).
det(P)
Important properties of similar matrices

Suppose A and B are similar matrices. Then the following

properties hold :

I A and B are equivalent.

I A and B have the same rank.

I det(B) = det(P −1 AP) = det(P −1 )det(A)det(P)

1
= det(A)det(P) = det(A).
det(P)

I Several other invariants of A and B are the same such as the

characteristic polynomial, minimal polynomial and eigen
values (with multiplicity).
An example of similar matrices
Consider the linear transformation f : R3 → R3 where
f (x, y , z) = (−x + y + z, x − y + z, x + y − z).
An example of similar matrices
Consider the linear transformation f : R3 → R3 where
f (x, y , z) = (−x + y + z, x − y + z, x + y − z).

Let β = γ both be the standard ordered basis of R3 .

An example of similar matrices
Consider the linear transformation f : R3 → R3 where
f (x, y , z) = (−x + y + z, x − y + z, x + y − z).

Let β = γ both be the standard ordered basis of R3 .

Then we get :

f (1, 0, 0) = (−1, 1, 1) = −1(1, 0, 0) + 1(0, 1, 0) + 1(0, 0, 1)

f (0, 1, 0) = (1, −1, 1) = 1(1, 0, 0) − 1(0, 1, 0) + 1(0, 0, 1)
f (0, 0, 1) = (1, 1, −1) = 1(1, 0, 0) + 1(0, 1, 0) − 1(0, 0, 1)
An example of similar matrices
Consider the linear transformation f : R3 → R3 where
f (x, y , z) = (−x + y + z, x − y + z, x + y − z).

Let β = γ both be the standard ordered basis of R3 .

Then we get :

f (1, 0, 0) = (−1, 1, 1) = −1(1, 0, 0) + 1(0, 1, 0) + 1(0, 0, 1)

f (0, 1, 0) = (1, −1, 1) = 1(1, 0, 0) − 1(0, 1, 0) + 1(0, 0, 1)
f (0, 0, 1) = (1, 1, −1) = 1(1, 0, 0) + 1(0, 1, 0) − 1(0, 0, 1)

 the matrixof f corresponding to the standard ordered basis

Hence
−1 1 1
is 1 −1 1 .

1 1 −1
Example (contd.)

Consider another ordered basis β 0 = (1, 1, 1), (−1, 1, 0), (−1, 0, 1).
Example (contd.)

Consider another ordered basis β 0 = (1, 1, 1), (−1, 1, 0), (−1, 0, 1).

Then we have the following:

f (1, 1, 1) = (1, 1, 1) = 1(1, 1, 1) + 0(−1, 1, 0) + 0(−1, 0, 1)

f (−1, 1, 0) = (2, −2, 0) = 0(1, 1, 1) − 2(−1, 1, 0) + 0(−1, 0, 1)
f (−1, 0, 1) = (2, 0, −2) = 0(1, 1, 1) + 0(−1, 1, 0) − 2(−1, 0, 1)
Example (contd.)

Consider another ordered basis β 0 = (1, 1, 1), (−1, 1, 0), (−1, 0, 1).

Then we have the following:

f (1, 1, 1) = (1, 1, 1) = 1(1, 1, 1) + 0(−1, 1, 0) + 0(−1, 0, 1)

f (−1, 1, 0) = (2, −2, 0) = 0(1, 1, 1) − 2(−1, 1, 0) + 0(−1, 0, 1)
f (−1, 0, 1) = (2, 0, −2) = 0(1, 1, 1) + 0(−1, 1, 0) − 2(−1, 0, 1)

Hence
 the matrix
 of f corresponding to the ordered basis β 0 is
1 0 0
0 −2 0 .
0 0 −2
Example (contd.)

 
1 −1 −1
Let P = 1 1 0 .
1 0 1
Example (contd.)

   
1 −1 −1 1/3 1/3 1/3
Let P = 1 1 0 . Then P −1 = −1/3 2/3 −1/3.
1 0 1 −1/3 −1/3 2/3
Example (contd.)

   
1 −1 −1 1/3 1/3 1/3
Let P = 1 1 0 . Then P −1 = −1/3 2/3 −1/3.
1 0 1 −1/3 −1/3 2/3
   
1/3 1/3 1/3 −1 1 1 1 −1 −1
Then P −1 AP = −1/3 2/3 −1/3  1 −1 1  1 1 0
−1/3 −1/3 2/3 1 1 −1 1 0 1
Example (contd.)

   
1 −1 −1 1/3 1/3 1/3
Let P = 1 1 0 . Then P −1 = −1/3 2/3 −1/3.
1 0 1 −1/3 −1/3 2/3
   
1/3 1/3 1/3 −1 1 1 1 −1 −1
Then P −1 AP = −1/3 2/3 −1/3  1 −1 1  1 1 0
−1/3 −1/3 2/3 1 1 −1 1 0 1
    
1/3 1/3 1/3 1 −1 −1 1 0 0
= 2/3
 −4/3 2/3  1 1 0  = 0 −2 0 .
2/3 2/3 −4/3 1 0 1 0 0 −2
Example (contd.)

   
1 −1 −1 1/3 1/3 1/3
Let P = 1 1 0 . Then P −1 = −1/3 2/3 −1/3.
1 0 1 −1/3 −1/3 2/3
   
1/3 1/3 1/3 −1 1 1 1 −1 −1
Then P −1 AP = −1/3 2/3 −1/3  1 −1 1  1 1 0
−1/3 −1/3 2/3 1 1 −1 1 0 1
    
1/3 1/3 1/3 1 −1 −1 1 0 0
= 2/3
 −4/3 2/3  1 1 0  = 0 −2 0 .
2/3 2/3 −4/3 1 0 1 0 0 −2

Hence A and B are similar matrices.

Another example
Consider the linear transformation seen earlier :
Another example
Consider the linear transformation seen earlier :

f : R2 → R2
f (x, y ) = (2x, y )
Another example
Consider the linear transformation seen earlier :

f : R2 → R2
f (x, y ) = (2x, y )

Consider the ordered basis (1, 0), (1, 1) for R2 . Then we have the
following:

f (1, 0) = (2, 0) = 2(1, 0) + 0(1, 1)

f (1, 1) = (2, 1) = 1(1, 0) + 1(1, 1)
Another example
Consider the linear transformation seen earlier :

f : R2 → R2
f (x, y ) = (2x, y )

Consider the ordered basis (1, 0), (1, 1) for R2 . Then we have the
following:

f (1, 0) = (2, 0) = 2(1, 0) + 0(1, 1)

f (1, 1) = (2, 1) = 1(1, 0) + 1(1, 1)

Hence the matrix of f corresponding to this ordered basis is :

 
2 1
A=
0 1
Another example (contd.)

Consider the standard ordered basis (1, 0), (0, 1) for R2 .

Another example (contd.)

Consider the standard ordered basis (1, 0), (0, 1) for R2 .

Then we have the following :

f (1, 0) = (2, 0) = 2(1, 0) + 0(0, 1)

f (0, 1) = (0, 1) = 0(1, 0) + 1(0, 1)
Another example (contd.)

Consider the standard ordered basis (1, 0), (0, 1) for R2 .

Then we have the following :

f (1, 0) = (2, 0) = 2(1, 0) + 0(0, 1)

f (0, 1) = (0, 1) = 0(1, 0) + 1(0, 1)

Hence the matrix of f corresponding to this ordered basis is :

 
2 0
B=
0 1
Another example (contd.)

 
1 1
Let P = .
0 1
Another example (contd.)

   
1 1 −1 1 −1
Let P = . Then P = .
0 1 0 1
Another example (contd.)

   
1 1 −1 1 −1
Let P = . Then P = .
0 1 0 1
   
1 12 1 1 −1
Then P −1 AP =
0 10 1 0 1
    
2 2 1 −1 2 0
= = .
0 1 0 1 0 1
   
2 1 2 0
Hence the matrices and are similar.
0 1 0 1
Linear transformations and similarity of matrices

Consider a linear transformation T : V → V and two ordered bases

β and γ for V .
Linear transformations and similarity of matrices

Consider a linear transformation T : V → V and two ordered bases

β and γ for V .

Let A be the matrix corresponding to T with respect to the basis β

and
Linear transformations and similarity of matrices

Consider a linear transformation T : V → V and two ordered bases

β and γ for V .

Let A be the matrix corresponding to T with respect to the basis β

and B the matrix corresponding to T with respect to the basis γ.
Linear transformations and similarity of matrices

Consider a linear transformation T : V → V and two ordered bases

β and γ for V .

Let A be the matrix corresponding to T with respect to the basis β

and B the matrix corresponding to T with respect to the basis γ.

Then A is similar to B !
Linear transformations and similarity of matrices

Consider a linear transformation T : V → V and two ordered bases

β and γ for V .

Let A be the matrix corresponding to T with respect to the basis β

and B the matrix corresponding to T with respect to the basis γ.

Then A is similar to B !

Why do we care about similarity?

Linear transformations and similarity of matrices

Consider a linear transformation T : V → V and two ordered bases

β and γ for V .

Let A be the matrix corresponding to T with respect to the basis β

and B the matrix corresponding to T with respect to the basis γ.

Then A is similar to B !

Why do we care about similarity? Because under some basis, we

hope that the corresponding matrix is a diagonal matrix which gives
an easy geometric understanding of the linear transformation.
Thank you