Which vectors remain unchanged in direction after a transformation?
That is, for a matrix \(A\), what vectors \(v\) are equal to scalar multiplication by \(\lambda\) following the operation of the matrix.
\(Av=\lambda v\)
The spectrum of a matrix is the set of its eigenvalues.
If eigen vectors space space, we can write
\(v=\sum_i \alpha_i | \lambda_i\rangle\)
Under what circumstances do they span the entirity?
The characteristic polynomial of a matrix is a polynomial whose roots are the eigenvalues of the matrix.
We know from the definition of eigenvalues and eigenvectors that:
\(Av=\lambda v\)
Note that
\(Av-\lambda v=0\)
\(Av-\lambda Iv=0\)
\((A-\lambda I)v=0\)
Trivially we see that \(v=0\) is a solution.
Otherwise matrix \(A-\lambda I\) must be non-invertible. That is:
\(Det(A-\lambda I)=0\)
For example
\(A=\begin{bmatrix}2&1\\1 & 2\end{bmatrix}\)
\(A-\lambda I=\begin{bmatrix}2-\lambda &1\\1 & 2-\lambda \end{bmatrix}\)
\(Det(A-\lambda I)=(2-\lambda )(2-\lambda )-1\)
When this is \(0\).
\((2-\lambda )(2-\lambda )-1=0\)
\(\lambda =1,3\)
You can plug this into the original problem.
For example
\(Av=3v\)
\(\begin{bmatrix}2&1\\1 & 2\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}=3\begin{bmatrix}x_1\\x_2\end{bmatrix}\)
As vectors can be defined at any point on the line, we normalise \(x_1=1\).
\(\begin{bmatrix}2&1\\1 & 2\end{bmatrix}\begin{bmatrix}1\\x_2\end{bmatrix}=\begin{bmatrix}3\\3x_2\end{bmatrix}\)
Here \(x_2=1\) and so the eigenvector corresponding to eigenvalue \(3\) is:
\(\begin{bmatrix}1\\1\end{bmatrix}\)
The trace of a matrix is the sum of its diagonal components.
\(Tr(M)=\sum_i^nm_{ii}\)
The trace of a matrix is equal to the sum of its eigenvectors.
Traces can be shown as the sum of inner products.
\(Tr(M)=\sum_i^ne_iMe^i\)
Traces commute
\(Tr(AB)=Tr(BA)\)
Traces of \(1\times 1\) matrices are equal to their component.
\(Tr(M)=m_{11}\)
If we want to manipulate the scalar:
\(v^TMv\)
We can use properties of the trace.
\(v^TMv=Tr(v^TMv)\)
\(v^TMv=Tr([v^T][Mv])\)
\(v^TMv=Tr([Mv][v^T])\)
\(v^TMv=Tr(Mvv^T)\)
For a square matrix \(M\) we can calculate \(MMMM...\), or \(M^n\) where \(n\in \mathbb{N}\).
Generally, calculating a matrix to an integer power can be complicated. For diagonal matrices it is trivial.
For a diagonal matrix \(M=D^n\), \(m_{ij}=d_{ij}^n\).
The exponential of a complex number is defined as:
\(e^x=\sum \dfrac{1}{j!}x^j\)
We can extend this definition to matrices.
\(e^X:=\sum \dfrac{1}{j!}X^j\)
The dimension of a matrix and its exponential are the same.
If we have \(e^A=B\) where \(A\) and \(B\) are matrices then we can say that \(A\) is matrix logarithm of \(B\).
That is:
\(\log B=A\)
The dimensions of a matrix and its logarithm are the same.
For a matrix \(M\), the square root \(M^{\dfrac{1}{2}}\) is \(A\) where \(AA=M\).
This does not necessarily exist.
Square roots may not be unique.
Real matrices may have no real square root.
In hermitian, show all symmtric matrices are hermitian
For a diagonal matrix, eigenvalues are the diagonal entries?
Similar matrix:
\(M=P^{-1}AP\)
\(M\) and \(A\) have the same eigenvalues. If \(A\) diagonal, then entries are eigenvalues.
If matrix \(M\) is diagonalisable if there exists matrix \(P\) and diagonal matrix \(A\) such that:
\(M=P^{-1}AP\)
If these exist then we can more easily work out matrix powers.
\(M^n=(P^{-1}AP)^n=P^{-1}A^nP\)
\(A^n\) is easy to calculate, as each entry in the diagonal taken to the power of \(n\).
Defective matrices are those which cannot be diagonalised.
Non-singular matries can be defective or not defective, for example the identiy matrix.
Singular matrices can also be defective or not defective, for example the empty matrix.
Consider an eigenvector \(v\) and eigenvalue \(\lambda\) of matrix \(M\).
We known that \(Mv=\lambda v\).
If \(M\) is full rank then we can generalise for all eigenvectors and eigenvalues:
\(MQ=Q\Lambda\)
Where \(Q\) is the eigenvectors as columns, and \(\Lambda\) is a diagonal matrix with the corresponding eigenvalues. We can then show that:
\(M=Q\Lambda Q^{-1}\)
This is only possible to calculate if the matrix of eigenvectors is non-singular. Otherwise the matrix is defective.
If there are linearly dependent eigenvectors then we cannot use eigen-decomposition.
This can be used to invert \(M\).
We know that:
\(M^{-1}=(Q\Lambda Q^{-1})^{-1}\)
\(M^{-1}=Q^{-1}\Lambda^{-1}Q\)
We know \(\Lambda\) can be easily inverted by taking the reciprocal of each diagonal element. We already know both \(Q\) and its inverse from the decomposition.
If any eigenvalues are \(0\) then \(\Lambda\) cannot be inverted. These are singular matrices.
We define a function, the commuter, between two objects \(a\) and \(b\) as:
\([a,b]=ab-ba\)
For numbers, \(ab-ba=0\), however for matrices this is not generally true.
Consider two matrices which share an eigenvector \(v\).
\(Av=\lambda_A v\)
\(Bv=\lambda_B v\)
Now consider:
\(ABv=A\lambda_B v\)
\(ABv=\lambda_A\lambda_B v\)
\(BAv=\lambda_A\lambda_B v\)
If the matrices share all the same eigenvectors, then the matrices commute, and \(AB=BA\).
\(A=A^{mn}\)
\(B=B^{no}\)
\(C=C^{mo}=A.B\)
\(c_{ij}=\sum_{r=1}^na_{ir}b_{rj}\)
Matrix multiplication depends on the order. Unlike for real numbers,
\(AB\ne BA\)
Matrix multiplication is not defined unless the condition above on dimensions is met.
A matrix multiplied by the identity matrix returns the original matrix.
For matrix \(M=M^{mn}\)
\(M=MI^m=I^nM\)
\(2\) matricies of the same size, that is with idental dimensions, can be added together.
If we have \(2\) matrices \(A^{mn}\) and \(B^{mn}\)
\(C=A+B\)
\(c_{ij}=a_{ij}+b_{ij}\)
An empty matrix with \(0\)s of the same size as the other matrix is the identity matrix for addition.
A matrix can be multiplied by a scalar. Every element in the matrix is multiplied by this.
\(B=cA\)
\(b_{ij}=ca_{ij}\)
The scalar \(1\) is the identity scalar.
A matrix of dimensions \(m*n\) can be transformed into a matrix \(n*m\) by transposition.
\(B=A^T\)
\(b_{ij}=a{ji}\)
\((M^T)^T=M\)
\((AB)^T=B^TA^T\)
\((A+B)^T=A^T+B^T\)
\((zM)^T=zM^T\)
With conjugation we take the complex conjugate of each element.
\(B=\overline A\)
\(b_{ij}=\overline a_{ij}\)
\(\overline {(\overline A)}=A\)
\(\overline {(AB)}=(\overline A)( \overline B)\)
\(\overline {(A+B)}=\overline A+\overline B\)
\(\overline {(zM)}=\overline z \overline M\)
Like transposition, but with conjucate.
\(B=A^*\)
\(b_{ij}=\bar{a_{ji}}\)
Alternatively, and particularly in physics, the following symbol is often used instead.
\((A^*)^T=A^\dagger\)
The rank of a matrix is the dimension of the span of its component columns.
\(rank (M)=span(m_1,m_2,...,m_n)\)
The span of the rows is the same as the span of the columns.
A matrix where every element is \(0\). There is one for each dimension of matrix.
\(A=\begin{bmatrix}0& 0&...&0\\0 & 0&...&0\\...&...&...&...\\0&0&...&0\end{bmatrix}\)
A matrix where \(a_{ij}=0\) where \(i < j\) is upper triangular.
A matrix where \(a_{ij}=0\) where \(i > j\) is lower triangular.
A matrix which is either upper or lower triangular is a triangular matrix.
All symmetric matrices are square.
The identity matrix is an example.
A matrix where \(a_{ij}=a_{ji}\) is symmetric.
A matrix where \(a_ij=0\) where \(i\ne j\) is diagonal.
All diagonal matrices are symmetric.
The identity matrix is an example.
From invertible matrix section in endo
A matrix can only be inverted if it can be created from a combination of elementary row operations.
How can we identify if a matrix is invertible? We want to create a scalar from the matrix which tells us if this possible. We can this scalar the determinant.
For a matrix \(A\) we label the determinant \(|A|\), or \(\det A\)
We propose \(|A|=0\) when the matrix is not invertible.
So how can we identify the function we need to undertake on the matrix?
We know that linear dependence results in determinants of \(0\).
We can model this as a function on the columns of the matrix.
\(\det M = \det ([M_1, ...,M_n)\)
If there is linear depednence, for example if two columns are the same then:
\(\det ([M_1,...,M_i,...,M_i,...,M_n])=0\)
Similarly, if there is a column of \(0\) then the determinant is \(0\).
\(\det ([M_1,...,0,...,M_n])=0\)
Show linear in addition
How can we identify the determinant of less simple matrices? We can use the multilinear form.
\(\sum c_i\mathbf M_i=\mathbf 0\)
Where \(\mathbf c \ne \mathbf 0\)
Or:
\(M\mathbf c=\mathbf 0\)
A matrix can be shown in terms of its columns. \(A=[v_1,...,v_n]\)
\(\det A=\det [v_1,...,v_n]\)
\(\det A=\sum_{k_1=1}^m...\sum_{k_n=1}\prod_{i=1}^ma_{ik_i}\det ([e_{k_1},...,e_{k_n}])\)
If a whole row or columns is \(0\) then:
\(\det A=\det [v_1,...,v_i,...,v_n]\)
\(\det A'=\det [v_1,...,cv_i,...,v_n]\)
\(\det A=\det [v_1,...,v_i,...,v_n]\)
\(\det A'=\det [v_1,...,cv_i,...,v_n]\)
\(\det A'=c\det [v_1,...,v_i,...,v_n]\)
\(\det A'=c\det A\)
As a result, multiplying a column by \(0\) makes the determinant \(0\).
A matrix with a column of \(0\) therefore has determinant \(0\)
\(A=[a_1,...,a_i,...,a_i,...,a_n]\)
\(D(A)=D([a_1,...,a_i,...,a_i,...,a_n])\)
We know from Result 3 that swapping columns reverses the sign. Reversing columns results in the same matrix, so the determinant must be unchanged.
\(D(A)=-D(A)\)
\(D(A)=0\)
If a column is a linear combination of other columns, then the matrix cannot be inverted.
\(A=[a_1,...,\sum_{j\ne i}^{n}c_ja_j,...,a_n]\)
\(\det A=\det ([v_1,...,\sum_{j\ne i}^{n}c_jv_j,...,v_n])\)
\(\det A=\sum_{j\ne i}^{n}c_j\det ([v_1,...,v_j,...,v_n])\)
\(\det A=\sum_{j\ne i}^{n}c_j\det ([v_1,...,v_j,,...,v_j,...,v_n])\)
As there is a repeating vector:
\(\det A=0\)
\(A=[v_1,...,v_i+v_j,...,v_i+v_j,...,v_n]\)
We know.
\(\det A=0\)
\(\det A=\det ([a_1,...,a_i,...,a_i,...,a_n])+\det([a_1,...,a_i,...,a_j,...,a_n])+\det([a_1,...,a_j,...,a_i,...,a_n])+\det([a_1,...,a_j,...,a_j,...,a_n])\)
So:
\(\det ([a_1,...,a_i,...,a_i,...,a_n])+\det ([a_1,...,a_i,...,a_j,...,a_n])+\det([a_1,...,a_j,...,a_i,...,a_n])+\det([a_1,...,a_j,...,a_j,...,a_n])=0\)
As \(2\) of these have equal columns these are equal to \(0\).
\(\det ([a_1,...,a_i,...,a_j,...,a_n])+\det ([a_1,...,a_j,...,a_i,...,a_n])=0\)
\(\det ([a_1,...,a_i,...,a_j,...,a_n])=-\det ([a_1,...,a_j,...,a_i,...,a_n])\)
We have
\(\det A=\sum_{k_1=1}^m...\sum_{k_n=1}\prod_{i=1}^ma_{ik_i}\det ([e_{k_1},...,e_{k_n}])\)
So what is the value of the determinant here?
We know that the determinant of the identity matrix is \(1\).
We know that the determinant of a matrix with identical columns is \(0\).
We know that swapping columns multiplies the determinant by \(-1\).
Therefore the determinants where the values of \(k\) are not all unique are \(0\).
The determinants of the others are either \(-1\) or \(1\) depending on how many swaps are required to restore to the identity matrix.
This is also shown as the Leibni formula.
\(\det A = \sum_{\sigma \in S_n}sgn (\sigma )\prod_{i=1}^na_{i,\sigma_i}\)
\(\det I = 1\)
\(\det (AB)=\det A \det B\)
\(\det (M^{-1})=\dfrac{1}{\det M}\)
We know this because:
\(\det (MM^{-1})=\det I = 1\)
\(\det M \det M^{-1}=1\)
\(\det (M^{-1})=\dfrac{1}{\det M}\)
\(\det (M^*)=\overline {\det M}\)
\(\det (M^T)=\det M\)
\(\det (A+B)=\det A + \det B\)
\(\det cM = c^n\det M\)
The determinant is equal to the product of the eigenvalues.
\(M=\begin{bmatrix}a & b\\c & d\end{bmatrix}\)
\(|M|=ad-bc\)
\(M=\begin{bmatrix}a & b & c\\d & e & f\\g & h & i\end{bmatrix}\)
\(|M|=aei+bfg+cdh-ceg-dbi-afh\)