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2021 postgraduate entrance examination mathematics 2 linear algebra

2022-07-03 11:04:00 【hflag168】

2021 Graduate school entrance examination, number two linear algebra

One choice question ( Every little question 5 branch )

Quadratic type $f(x_1, x_2, x_3)=(x_1+x_2)^2 + (x_2+x_3)^2-(x_3-x_1)^2$ The positive inertia index and negative inertia index of are
A) 2, 0. B) 1,1. C) 2,1. D) 1,2.
Let third-order matrix $A=(\alpha_1, \alpha_2,\alpha_3)$ , $B=(\beta_1, \beta_2, \beta_3)$ , If the vector group $\alpha_1, \alpha_2, \alpha_3$ Can be $\beta_1, \beta_2$ The linear table out , be
A) Ax=0 All solutions of are Bx=0 Solution
B) $A^Tx=0$ All solutions of are $B^Tx=0$ Solution .
C) Bx=0 All solutions of are Ax=0 Solution .
D) $B^Tx=0$ All solutions of are $A^Tx=0$ Solution .
Known matrix $A=\begin{pmatrix} 1 & 0 &-1 \\ 2 & -1 & 1\\ -1 & 2 & -5 \end{pmatrix}$ If lower triangular invertible matrix P And upper triangular invertible matrix Q, send PAQ Is the focus matrix , be P, Q You can take... Separately
A) $\begin{pmatrix}1&0&0\\ 0&1&0\\0&0&1 \end{pmatrix}$ , $\begin{pmatrix} 1 & 0&1\\ 0& 1&3\\0&0&1\end{pmatrix}$ . B) $\begin{pmatrix}1&0&0\\ 2& -1& 0\\ -3&2&1 \end{pmatrix}$ , $\begin{pmatrix} 1&0&0\\ 0&1&0\\0&0&1\end{pmatrix}$ .
C) $\begin{pmatrix} 1&0&0\\2&-1&0\\-3&2&1\end{pmatrix}$ , $\begin{pmatrix}1&0&1\\0&1&3\\0&0&1 \end{pmatrix}$ . D) $\begin{pmatrix} 1&0&0\\0&1&0\\ 1&3&1\end{pmatrix}$ , $\begin{pmatrix} 1&2 &-3\\0&-1&2\\0&0&1\end{pmatrix}$ .

Two Completion

polynomial $f(x)=\begin{vmatrix}x&x&1&2x\\1&x&2&-1\\2&1&x&1\\2&-1&1&x \end{vmatrix}$ in $x^3$ The coefficient of the term is ___________________________.(6 branch )

3、 ... and Answer the question

Let's set the matrix $A=\begin{pmatrix}2&1&0\\1&2&0\\1&a&b \end{pmatrix}$ There are only two different eigenvalues . if A Similar to diagonal matrix , seek $a, b$ Value , And find the reversible matrix P, bring $P^{-1}AP$ It's a diagonal matrix .(12 branch )

analysis : This question mainly examines candidates' mastery of the theory of similar diagonalization of general matrices . The knowledge points involved include :

Eigenvalues and eigenvectors ( Basic knowledge of )
General matrix similar diagonalization theory : If all eigenvalues of a matrix are different , Then it must be similar diagonalization ; When a matrix exists k When the eigenvalue is repeated , as long as k Multiple eigenvalues correspond to k A linear independent eigenvector , Then the matrix can still be similarly diagonalized .

answer :

$\begin{vmatrix} 2-\lambda&1&0\\ 1& 2-\lambda&0\\1&a&b-\lambda\end{vmatrix}$ = $(b-\lambda)(\lambda-1)(\lambda-3)$

Because the matrix A There are only two different eigenvalues , So when 1 When it is a multiple eigenvalue , $\lambda_1=\lambda_2=1, \lambda_3=3$ , be b=1. here

$(A-E)=\begin{pmatrix}1&1&0\\1&1&0\\1&a&1 \end{pmatrix}\rightarrow\begin{pmatrix}1&1&0\\1&a&1\\0&0&0 \end{pmatrix}$ $\rightarrow \begin{pmatrix}1&1&0\\ 0&a-1&1\\0&0&0\end{pmatrix}$

be r(A-E)=2, This means matrix A The double eigenvalue of is 1 when , Only a linear independent eigenvector , matrix A Cannot be similar to diagonal matrix . Conflict with known .

Then you can choose $\lambda_1=1, \lambda_2=\lambda_3=3$ , be b=3. here

$(A-3E)=\begin{pmatrix}-1&1&0\\1&-1&0\\1&a&0\end{pmatrix}$ $\rightarrow \begin{pmatrix} 1&-1&0\\ 1&a&0\\0&0&0\end{pmatrix}$ $\rightarrow \begin{pmatrix} 1&-1&0\\ 0&a+1&0\\0&0&0\end{pmatrix}$

Then when $r (A - 3 E) = 1$ when , Double eigenvalue 3 It corresponds to two linear independent eigenvectors , here $a = - 1$ .

So you can get : $a = - 1, b = 3$ . Now the matrix $A=\begin{pmatrix}2&1&0\\1&2&0\\1&-1&3\end{pmatrix}$

When $\lambda_1=1$ when , The corresponding eigenvector is obtained as $\zeta_1=(-1, 1, 1)^T$

When $\lambda_2=\lambda_3=3$ when , The corresponding linear independent eigenvector is obtained as $\zeta_2=(1,1,0)^T, \zeta_3=(0,0,1)^T$

Make $P=(\zeta_1, \zeta_2, \zeta_3)=\begin{pmatrix} -1&1&0\\1&1&0\\1&0&1\end{pmatrix}$ Then there are $P^{-1}AP=\begin{pmatrix} 1&&\\&3&\\&&3\end{pmatrix}$ For diagonal matrix .