Matrix
A and A' are equivalent
A and A' are similar
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A′m×n=T−1m×m⋅Am×nSn×n det(T),det(S)≠0
m==n, T=S
square matrix
det(A) =0?
No
singular
invertible
Yes
\[det(A^{-1})=1/det(A)\]
vectors in A are linearly independent
\[A_{n \times n}x=\lambda x\]
\[E\lambda\] eigenspace of A: the set of all eigenvectors {x}, eigenvectors associated with different eigenvalues are linearly independent, A and At process the same eigenvalues but not same vectors
the spectrum of A= the set of all eigenvalues ofA
\[p(\lambda) = det(A-\lambda I) = (−1)^n (λ − c1) ^{r _1} · · · (λ − c_k ) ^{r_k}\] characteristic polynomial of A,
\[=0, \lambda\] is an eigenvalue, and the root of the p.
A' is diagnolisable
iff A is a diagonal matrix
\[dim(Im(A'-c_iI_n)) = n-r_i\] \[dim(ker(A'-c_iI_n))=r_i\] ci is a different root.
\[A'= D=S^{-1}AS, S=[e_1,e_2,e_3,...,e_n]\]
Constructing the diagonal form of A' using eigenvectors
applications
differential applications for differential equations.
p(A) =0, cayley-hamilton theorem, useful for find the inverse of a matrix (can be expressessed as a linear combination of its powers)
Vectors
scalar/dot product/inner product\[ u \cdot v =||u|| ||v|| cos \theta = \sum_{i=1}^{n}a_i b_i =a^T b = b^T a\]
Triangular inequality \[ ||x+y|| \leq ||x||+||y||\]
Cauchy-Schwarz inequality\[|x \cdot y| \leq ||x|| ||y|| \]
Parallelogram law\[||x+y|| + ||x-y||=||x||^2+||y||^2\]
\[P^2=P, P_\pi x=p\]
projection matrix, eigenvalues are either 0 or 1, \[P=\frac{bb^T}{||b|||^2 (i.e.b^Tb)}\], b basis bector
normal equation:\[B_{n \times m}^TB\lambda =B^T x\] \[p=B \lambda =B(B^TB)^{-1}B^Tx\]
\[x \cdot y=0 \]
orthogonal
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orthogonal projection when \[P^T=P\]