2020年北京大学高等代数与解析几何考研试题
1.(10分)设$V_0={0},V_1,cdots,V_{n-1},V_{n}={0}$
是$n+1$个有限维线性空间,定义线性变换$varphi_i:V_{i} o V_{i+1},i=0,1,2,cdots,n-1$,若对$i =0,1,2,cdots,n-1$均有$mathrm{ker} varphi_{i+1} = mathrm{Im}varphi_i$中,证明$displaystylesum_{i=0}^n(-1)^imathrm{dim}(V_i)=0$.
2. (15分)设$c_0,c_1,cdots,c_k$是$k+1$个复数,证明:存在唯一一个次数不超过$k$的复系数多项式函数$p(x)$使得$p(0)=c_0,p(1)=c_1,cdots,p(k)=c_k$,且这样的多项式是唯一的.
3. (20分)设$A$是秩为$r$的实对称矩阵,试证明必存在一个非零的$r$阶主子式使得它的行列式非零,并且任意一个非零的$r$阶主子式符号相同.
4. (20分)设$n$阶方阵$A=(a_{ij})_{n imes n}$可相似对角化,它的特征值为$lambda_1,cdots,lambda_n$,每个特征值$lambda_i$的特征子空间都由一族特征向量$alpha_ {ij_1},cdots,alpha_{ij_n}$张成,设$A^ast=(A_{ji})_{n imes n}$, $A_{ji}$是$a_{ji}$对应的代数余子式,求$A^ast$的特征值和特征向量.
5. (15分)设$varphi$是一个线性变换, $lambda_1,cdots,lambda_k$是$varphi$的特征值,证明$varphi$可对角化的充分必要条件是对$varphi$的每个特征值$lambda$,均有$mathrm{dim}left(mathrm{Im} (lambda mathrm{id}-varphi) ight)=mathrm{dim}left(mathrm{Im} (lambda mathrm{id}-varphi)^2 ight)$,其中$mathrm{id}$是恒等变换.
6. (15分)设$eta$是欧氏空间$V$中的单位向量,定义镜像变换$sigma:sigma(alpha)=alpha-2(alpha,eta)eta$,其中$( , )$表示內积.
(1) 证明$sigma$是正交变换.
(2) 证明$V$的任意正交变换都可以表示成若干镜像变换的乘积.
7. (15分)已知向量$overrightarrow{u},overrightarrow{v},overrightarrow{w}$满足$left|overrightarrow{u} ight|=left|overrightarrow{v} ight|=left|overrightarrow{w} ight|> 0,overrightarrow{u}cdot overrightarrow{v}=overrightarrow{v}cdot overrightarrow{w}=overrightarrow{w}cdot overrightarrow{u}$,若对任意非零向量$overrightarrow{x}$,均存在实数$a,b,c$,使得$overrightarrow{x} imes overrightarrow{u}=aoverrightarrow{u} +boverrightarrow{v}+coverrightarrow{w},overrightarrow{x} imes overrightarrow{v}=aoverrightarrow{v}+ boverrightarrow{w}+coverrightarrow{u}$,证明$overrightarrow{x} imes overrightarrow{w}=aoverrightarrow{w}+ boverrightarrow{u} +coverrightarrow{v}$.
8. (20分)设平面直角坐标系下二次曲线$gamma$的方程为$x^2 + 2y^2 + 6xy + 8x+ 10y +6=0$.
(1) 证明$gamma$是双曲线.
(2) 求$gamma$的长半轴,短半轴的方程与长轴和短轴长,并且说明哪条与$gamma$相交.
9. (20分)求椭圆$x^2 +8y^2 +4xy + 10x + 12y+4=0$的内接三角形的面积的最大值.
2020年北京大学数学分析考研试题
1. (15分)定义在$[a,b]$上的函数$f(x)$满足:任取$x_0in [a,b]$,均有$limsup_{x o x_0}f(x)leq f(x_0)$,问$f(x)$在$[a,b]$上是否有最大值,给出证明或反例.
2. (15分)判断$displaystyle f(x)=frac{x}{1+xcos^2 x}$在$[0,+infty)$上是否一致连续,并说明理由.
3. (15分) $f(x)$在$[1,+infty)$连续且满足:对任意$x,yin [1,+infty)$,有$f(x+y)leq f(x)+f(y)$.问$displaystylelim_{x o+infty}frac{f(x)}{x}$是否存在,给出证明或反例.
4. (15分,第一小题7分,第二小题8分)已知$f(x)$在$[0,1]$连续,单调增加且$f(x)geq 0$,记
$$s=frac{int_{0}^{1}xf(x)\,mathrm{d}x}{int_{0}^{1}f(x)\,mathrm{d}x}.$$
(1)证明$sgeq frac{1}{2}$.
(2)比较$displaystyleint_{0}^{s}f(x)\,mathrm{d}x$与$displaystyleint_{s}^{1}f(x)\,mathrm{d}x$的大小. (可以用物理或几何直觉)
5. (15分)根据$displaystyleint_{0}^{+infty}frac{sin x}{x}\,mathrm{d}x=frac{pi}{2}$,计算$displaystyleint_{0}^{+infty}left(frac{sin x}{x} ight)^2\,mathrm{d}x$,并说明计算依据.
6. (15分)在承认平面Green公式的前提下证明如下特殊情况下的Stokes公式
$$oint_Gamma R(x,y,z)\,mathrm{d}z=iint_Sigmafrac{partial R}{partial y}dydz-frac{partial R}{partial x}dzdx.$$
7. (20分,第一小题10分,第二小题10分) (1)设$0< p<1$,求$f(x)=cos px$在$[-pi,pi]$上的Fourier级数.
(2)证明余元公式
$$int_{0}^{1}x^{p-1}(1-x)^{-p}dx=frac{pi}{sin(ppi)}.$$
8. (20分)设$C_r$为半径为$r$的圆周, $f(x,y)$满足$displaystyle f(0,0)=0,frac{partial^2f}{partial x^2}+frac{partial^2f}{partial y^2}=x^2+y^2$, $f(x,y)$是$C^2$的,计算$displaystyle A(r)=int_{C_r}f(x,y)\,mathrm{d}s$.
9. (20分,第一小题12分,第二小题8分)设$q_kgeq p_k>0$, $q_{k+1}-q_kgeq p_k+p_{k+1}$且$displaystylesum_{k=1}^{infty}a_kln p_k=+infty$,记
$$T_{p_k,q_k}(x) riangleq frac{cos (q_k+p_k)x}{p_k}+frac{cos (q_k+p_k-1)x}{p_k-1}+frac{cos (q_k+p_k-2)x}{p_k-2}+cdots+frac{cos (q_k+1)x}{1}$$
$$ -frac{cos (q_k-1)x}{1}-frac{cos (q_k-2)x}{2}-cdots-frac{cos (q_k-p_k)x}{p_k},$$
设$displaystyle a_kgeq 0,sum_{k=1}^{infty}a_k<+infty$, $displaystyle f(x)=sum_{k=1}^{infty}a_kT_{p_k,q_k}(x)$.
(1) 求证: $f(x)$是在$mathbb{R}$上连续的以$2pi$为周期的周期函数.
(2) 判断并证明: $f(x)$的Fourier级数在$x=0$处的收敛性.
设$overrightarrow{x} imes overrightarrow{w}=Aoverrightarrow{w}+ Boverrightarrow{u} +Coverrightarrow{v}$, $left|overrightarrow{u}
ight|=left|overrightarrow{v}
ight|
=left|overrightarrow{w}
ight|=m,overrightarrow{u}cdot overrightarrow{v}=overrightarrow{v}cdot overrightarrow{w}=overrightarrow{w}cdot overrightarrow{u}=n$.
利用混合积的性质可知
[(overrightarrow{u},overrightarrow{x},overrightarrow{v})=
overrightarrow{u}cdot (overrightarrow{x} imes overrightarrow{v})=(overrightarrow{u} imes overrightarrow{x})cdot overrightarrow{v},]
于是
[aoverrightarrow{u}cdot overrightarrow{v}+boverrightarrow{u}cdot overrightarrow{w}
+cleft|overrightarrow{u}
ight|^2=-left(aoverrightarrow{u}cdot overrightarrow{v}+bleft|overrightarrow{u}
ight|^2
ight)+coverrightarrow{v}cdot overrightarrow{w}.]
即
[(2a+b+c)m+(b+c)n=0.]
又
[(overrightarrow{u},overrightarrow{x},overrightarrow{w})=
overrightarrow{u}cdot (overrightarrow{x} imes overrightarrow{w})=(overrightarrow{u} imes overrightarrow{x})cdot overrightarrow{w},]
于是
[Aoverrightarrow{u}cdot overrightarrow{w}+Bleft|overrightarrow{u}
ight|^2+coverrightarrow{u}cdot overrightarrow{v}
=-left(aoverrightarrow{u}cdot overrightarrow{w}+boverrightarrow{v}cdot overrightarrow{w}+cleft|overrightarrow{w}
ight|^2
ight).]
即
[(A+C+a+b)m+(b+c)n=0.]
由
[(overrightarrow{v},overrightarrow{x},overrightarrow{w})=
overrightarrow{v}cdot (overrightarrow{x} imes overrightarrow{w})=(overrightarrow{v} imes overrightarrow{x})cdot overrightarrow{w},]
于是
[Aoverrightarrow{v}cdot overrightarrow{w}+Boverrightarrow{u}cdot overrightarrow{v}+Cleft|overrightarrow{v}
ight|^2
=-left(aoverrightarrow{v}cdot overrightarrow{w}+bleft|overrightarrow{w}
ight|^2+coverrightarrow{u}cdot overrightarrow{w}
ight).]
即
[(A+B+a+c)m+(C+b)n=0.]