### Matrix calculus: Minimal polynomial

### Cayley-Hamilton

Let #A# be an \((n\times n)\)-matrix. *Earlier*, we saw that all matrices of this size, with the usual matrix addition and scalar multiplication, form an #n^2#-dimensional vector space. Thus, there is a non-negative integer #k\le n^2#, such that the system \[\basis{1,A,A^2,\ldots, A^{k}}\] is linearly dependent. In that case there is an equality of the form

\[A^k+c_{k-1} \cdot A^{k-1}+\cdots +c_1\cdot A + c_0\cdot I_n = 0\]

According to the Cayley-Hamilton theorem below, #k# is at most #n#. To understand this result, we specify what is meant by evaluating a polynomial at a square matrix.

Let #n# be a natural number and #A# an #(n\times n)#-matrix. Under **substitution **of #A# (for #x#) into a polynomial \[ p(x) = c_0+c_1x+\cdots+ c_kx^k\] in #x#, or **evaluating** this polynomial at #A#, we mean determining the #(n\times n)#-matrix \(p(A)\) defined by \[p(A)= c_0\cdot I_n+c_1\cdot A+\cdots +c_k\cdot A^k\]

The assignment of #p(A)# to #p(x)# is a mapping # P\to M_{n\times n}# from the vector space #P# of all polynomials in #x# to the vector space #M_{n\times n}# of all #(n\times n)#-matrices.

This map respects multiplication in the following sense: if #p(x)#, #q(x)#, and #r(x)# are polynomials such that #r (x)= p(x)\cdot q(x)#, then \[r(A) = p(A)\, q(A)\] Under substitution of a linear map #L: V\to V# into the polynomial #p(x)# as given above, we mean the linear map #p(L): V\to V# defined by \[p(L)= c_0\cdot I_V+c_1\cdot L+\cdots +c_k\cdot L^k\]

A remarkable fact about linear maps from a finite-dimensional vector space to itself is that substitution in their *characteristic polynomial* gives the *null map*:

Cayley-Hamilton Each linear map #L:V\to V# from a finite-dimensional vector space #V# to itself satisfies \[p_L(L) = 0_V\]where #p_L(x)# is the characteristic polynomial.

In order to calculate the matrix #p(A)#, we substitute #A# for #x# in the polynomial #p(x)# and we simplify the result to a single #(3\times3)#-matrix:

\[\begin{array}{rcl}

p(A) &=&A^2-3\cdot A-4\cdot I_3\\

&&\phantom{xxx}\color{blue}{A\text{ substituted }}\\

&=& \matrix{1 & 0 & 0 \\ -15 & 16 & 0 \\ 30 & -30 & 1 \\ } -3\cdot \matrix{1 & 0 & 0 \\ -3 & 4 & 0 \\ 8 & -10 & -1 \\ } -4\cdot \matrix{1&0&0\\ 0&1&0\\ 0&0&1}\\

&&\phantom{xxx}\color{blue}{\text{explicit matrices substituted}}\\

&=& \matrix{-6 & 0 & 0 \\ -6 & 0 & 0 \\ 6 & 0 & 0 \\ }\\

&&\phantom{xxx}\color{blue}{\text{linear combination simplified}}\\

\end{array}

\]

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