一道线性代数题目,3小问1 ,1-h^2-1 ,-1上面这个是个m的2*2矩阵.(1) For general h,diagonalize m.(2) Find the projectors to the eigenvalues of m (3) What happens with eigenvectors when h = What happens with m when h =
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一道线性代数题目,3小问1 ,1-h^2-1 ,-1上面这个是个m的2*2矩阵.(1) For general h,diagonalize m.(2) Find the projectors to the eigenvalues of m (3) What happens with eigenvectors when h = What happens with m when h =
一道线性代数题目,3小问
1 ,1-h^2
-1 ,-1
上面这个是个m的2*2矩阵.
(1) For general h,diagonalize m.
(2) Find the projectors to the eigenvalues of m
(3) What happens with eigenvectors when h = What happens with m when h =
一道线性代数题目,3小问1 ,1-h^2-1 ,-1上面这个是个m的2*2矩阵.(1) For general h,diagonalize m.(2) Find the projectors to the eigenvalues of m (3) What happens with eigenvectors when h = What happens with m when h =
1L的答案貌似跳了好多步骤,我教你吧- -你上百度,我加你
1。将矩阵对角化
|λE-m|=0
λ-1 h^2-1
1 λ+1 =0
即：(λ-1)(λ+1)-(h^2-1)=0
λ^2-h^2=0
λ=-h,h
则对角矩阵为
-h 0
0 h
2. 特征向量为:
当λ=-h,
(λE-m)p=0
-h-1 h^2-1 * p1 =0...
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1。将矩阵对角化
|λE-m|=0
λ-1 h^2-1
1 λ+1 =0
即：(λ-1)(λ+1)-(h^2-1)=0
λ^2-h^2=0
λ=-h,h
则对角矩阵为
-h 0
0 h
2. 特征向量为:
当λ=-h,
(λE-m)p=0
-h-1 h^2-1 * p1 =0
1 -h+1
对于系数矩阵
-h-1 h^2-1
1 -h+1
第2行乘以(h+1)加到第一行
0 0
1 -h+1
则p=(h-1,1)
当λ=h,
(λE-m)p=0
h-1 h^2-1 * p2 =0
1 h+1
对于系数矩阵
h-1 h^2-1
1 h+1
第2行乘以(1-h)加到第一行
0 0
1 h+1
则p=(h+1,-1)
则特征向量为：(p1,p2)
h-1 h+1
1 -1
h=0,则两个特征值相等，对角阵为零矩阵
h=0,此时，两个特征向量
p1=(-1,1)T (T为转置）
p2=(1,-1)T
显然，p1,p2线性相关，所以，对于特征值为0，两个特征向量线性相关，
则：m不可以对角化
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