What are eigenvalues ​​and eigenvectors 4

content
“Preliminary remark
" The definition
»The characteristic polynomial and the algebraic multiplicity
»Eigenvectors and the geometric multiplicity
“Examples

Preliminary remark

Eigenvalues ​​and eigenvectors are one of the most important topics in linear algebra. Their applications are very diverse. The idea is a simple one, we are looking for vectors \ (\ vec v_i \) which are mapped to a multiple of themselves by the matrix \ (A \), in formulas that means
\ begin {align *}
A \ vec {v} _i = \ lambda_i \ vec {v} _i.
\ end {align *}

The definition

Let us formalize the above definition even further. As so often, we use the letters \ (u, v, w \) for vectors instead of \ (\ vec {v}, \ vec {u}, \ vec {w} \). Matrices are named with capital letters \ (A, B, \ dots \) ​​and scalars with \ (\ lambda, \ alpha \). In addition, we only deal with matrices over the real field \ (\ mathbb {R} \). We consider the \ ((n \ times n) \) - matrix \ (A \) as a linear mapping over the vector space \ (\ mathbb {R} ^ n \). If a vector \ (v \ in V \) satisfies the equation
\ begin {align *}
Av = \ lambda v
\ end {align *}
for a \ (\ lambda \ in \ mathbb {R} \), we call a \ (v \) eigenvector with associated eigenvalue \ (\ lambda \).

A trivial example would be the matrix
\ begin {align *}
A = \ begin {pmatrix}
1 & 0\\
0 & 1
\ end {pmatrix}
\ end {align *}
with the associated eigenvectors \ (v_1 = (1; 0) \) and \ (v_2 = (0; 1) \), both have the eigenvalue \ (\ lambda_1 = 1 \). The matrix \ (A \) with
\ begin {align *}
\ left (
\ begin {array} {ccc}
0 & \ frac {1} {2} & \ frac {1} {2} \
\ frac {1} {2} & 0 & \ frac {1} {2} \
\ frac {1} {2} & \ frac {1} {2} & 0 \
\ end {array}
\ right)
\ end {align *}
has the eigenvectors
\ begin {align *}
v_1 = \ begin {pmatrix}
1\\1\\1
\ end {pmatrix}, v_2 = \ begin {pmatrix}
-1\\0\\1
\ end {pmatrix},
v_3 = \ begin {pmatrix}
-1\\1\\0
\ end {pmatrix}
\ end {align *}
with the eigenvalues
\ begin {align *}
\ lambda_1 = 1, \ lambda_2 = \ lambda_3 = - \ frac {1} {2}.
\ end {align *}
One can easily recalculate this, but how did we find the vectors and the eigenvalues?

The characteristic polynomial and algebraic multiplicity

We derive the characteristic polynomial so that eigenvalues ​​can be calculated. The train of thought is a simple one, fulfilling eigenvalue and eigenvector
\ begin {align *}
& Av = \ lambda v \
& Av- \ lambda v = 0 \
& (A- \ lambda I_n) v = 0,
\ end {align *}

where \ (I_n \) is the \ ((n \ times n) \) - identity matrix. Now the linear mapping / matrix \ ((A- \ lambda I_n) \) maps the vector \ (v \) to the zero vector. We therefore know that \ ((A- \ lambda I_n) \) cannot be injective, or that it is more important for us and does not have full rank. If the matrix \ ((A- \ lambda I_n) \) does not have a full rank, we know (LINK) that its determinant is 0. So we get
\ begin {align *}
\ det (A- \ lambda I_n) = 0.
\ end {align *}
For example, for the above matrix we calculate:
\ begin {align *}
\ det (
\ begin {pmatrix}
0 & \ frac {1} {2} & \ frac {1} {2} \
\ frac {1} {2} & 0 & \ frac {1} {2} \
\ frac {1} {2} & \ frac {1} {2} & 0 \
\ end {pmatrix}
- \ lambda
\ begin {pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\ end {pmatrix}) = 0 \
\ rightarrow \ det (\ begin {pmatrix}
- \ lambda & \ frac {1} {2} & \ frac {1} {2} \
\ frac {1} {2} & - \ lambda & \ frac {1} {2} \
\ frac {1} {2} & \ frac {1} {2} & - \ lambda \
\ end {pmatrix} = 0 \
- \ frac {1} {4} (\ lambda-1) (2 \ lambda + 1) ^ 2 = 0
\ end {align *}




This equation can be solved and has the solutions \ (\ lambda_1 = 1 \) and \ (\ lambda_ {2,3} = - \ frac {1} {2} \), these are our eigenvalues. We therefore get eigenvalues ​​by solving the so-called characteristic polynomial under the equation \ (\ det (A- \ lambda I_n) = 0 \).

It is said that \ (\ lambda_1 = 1 \) has algebraic multiplicity 1 and \ (\ lambda_ {2,3} = - \ frac {1} {2} \) has algebraic multiplicity 2, because it is a double solution.

Eigenvectors and geometric multiplicity

We just learned how to calculate eigenvalues. If we have this, the loosening of the associated one leads to it
\ begin {align *}
(A- \ lambda I_n) v = 0
\ end {align *}
after \ (v \). We already know that this system must always have a solution, because rank (\ ((A- \ lambda I_n) \)) = rank (\ ((A- \ lambda I_n) \) | 0). Let's get started and insert \ (\ lambda_1 \) to calculate \ (v_1 \)
\ begin {align *}
\ begin {pmatrix}
-1 & \ frac {1} {2} & \ frac {1} {2} \
\ frac {1} {2} & -1 & \ frac {1} {2} \
\ frac {1} {2} & \ frac {1} {2} & -1 \
\ end {pmatrix} v = 0
\ end {align *}

This system of equations (here (LINK) more about how we solve it) has the solution
\ begin {align *}
L = \ {s \ begin {pmatrix}
1\\1\\1
\ end {pmatrix} | s \ in \ mathbb {R} \}
\ end {align *}

and thereby we get the eigenvector \ (v = (1; 1; 1) \). The dimension of the solution is 1, we call this dimension the geometric multiplicity. The eigenvector (s) for the eigenvalue \ (\ lambda_ {2,3} = - \ frac {1} {2} \) (with algebraic multiplicity 2) are obtained by solving
\ begin {align *}
\ begin {pmatrix}
\ frac {1} {2} & \ frac {1} {2} & \ frac {1} {2} \
\ frac {1} {2} & \ frac {1} {2} & \ frac {1} {2} \
\ frac {1} {2} & \ frac {1} {2} & \ frac {1} {2} \
\ end {pmatrix} v = 0.
\ end {align *}

The matrix of this system of equations obviously has rank 1, therefore the solution of the system of equations has rank 2. It is given by
\ begin {align *}
L = \ {s \ begin {pmatrix}
-1\\0\\1
\ end {pmatrix} + r \ begin {pmatrix}
-1\\1\\0
\ end {pmatrix} | s, r \ in \ mathbb {R} \}
\ end {align *}

and this gives us the two eigenvectors \ (v_2 = (- 1; 0; 1) \) and \ (v_3 = (- 1; 1; 0) \). The eigenvalue \ (\ lambda_ {2,3} = - \ frac {1} {2} \) therefore has geometric multiplicity 2. In the examples you will find matrices whose eigenvalues ​​have a greater algebraic than geometric multiplicity.

Examples

Upper triangular matrix: Calculate eigenvalues ​​and eigenvectors of the matrix \ (B \) with
\ begin {align *}
B = \ left (
\ begin {array} {ccc}
1 & 2 & 3 \\
0 & 4 & 5 \\
0 & 0 & 6 \\
\ end {array}
\ right).
\ end {align *}
What do you notice?

Solution: It is an upper triangular matrix. Thereby we very easily get the characteristic polynomial by expanding after the first column
\ begin {align *}
\ det (B- \ lambda I_3) = - (\ lambda-6) (\ lambda-4) (\ lambda-1).
\ end {align *}

We recognize that with an upper triangular matrix we can apparently read off the eigenvalues ​​in the diagonal, because
\ begin {align *}
- (\ lambda-6) (\ lambda-4) (\ lambda-1) = 0
\ end {align *}

has, according to the product zero theorem, the solutions \ (\ lambda_1 = 6, \ lambda_2 = 4, \ lambda_3 = 1 \). We calculate the associated eigenvectors by solving the three systems of equations
\ begin {align *}
(B-6I_3) v = 0, (B-4I_3) v = 0, (B-1I_3) v = 0
\ end {align *}
and get the three eigenvectors \ (v_1 = (16; 25; 10) \), \ (v_2 = (2,3,0) \) and \ (v_3 = (1,0,0) \).

Geometric multiplicity: Upper triangular matrix: Calculate eigenvalues ​​and eigenvectors of the matrix \ (B \) with
\ begin {align *}
B = \ left (
\ begin {array} {ccc}
4 & 1 & 0 \\
0 & 4 & 0 \\
0 & 0 & 1 \\
\ end {array}
\ right).
\ end {align *}

What do you notice?

Solution: As explained in the previous example, we can read off the eigenvalues, \ (\ lambda_ {1,2} = 4 \) and \ (\ lambda_3 = 1 \) apply. \ (V_3 \) can easily be calculated as \ (\ lambda_3 \), it gives \ (v_3 = (0; 0; 1) \). Interesting is the eigenvalue \ (\ lambda_ {1,2} = 4 \) with algebraic multiplicity 2 then

\ begin {align *}
& (B-4I_3) v = 0 \ \
& \ left (
\ begin {array} {ccc}
0 & 1 & 0 \\
0 & 0 & 0 \\
0 & 0 & -3 \\
\ end {array}
\ right) v = 0
\ end {align *}


has only a one-dimensional solution, because \ ((B-4I_3) \) has rank 2. This solution is given by
\ begin {align *}
L = \ {s \ begin {pmatrix}
1\\0\\0
\ end {pmatrix} | s \ in \ mathbb {R} \}
\ end {align *}

and therefore we have only one eigenvector, \ (v_1 = (1; 0; 0) \). The eigenvalue \ (\ lambda_ {1,2} = 4 \) therefore has algebraic multiplicity 2 but only geometric multiplicity 1.

\ (2 \ times 2 \): Compute eigenvalues ​​and eigenvectors of the matrix \ (B \) with
\ begin {align *}
B = \ left (
\ begin {array} {cc}
1 & 2 \\
3 & 4 \\
\ end {array}
\ right).
\ end {align *}
The characteristic polynomial results in \ (\ lambda ^ 2-5 \ lambda -2 \), which has the rounded solutions \ (\ lambda_1 = 5.372 \) and \ (\ lambda_2 = -0.372 \). In general, eigenvalues ​​are of course not integers, even if we did this in our previous examples for the sake of simplicity. The associated eigenvectors are \ (v_1 = (0.457; 1) \) and \ (v_2 = (-1.457, 1) \).