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 =](/uploads/image/z/14333726-38-6.jpg?t=%E4%B8%80%E9%81%93%E7%BA%BF%E6%80%A7%E4%BB%A3%E6%95%B0%E9%A2%98%E7%9B%AE%2C3%E5%B0%8F%E9%97%AE1+%2C1-h%5E2-1+%2C-1%E4%B8%8A%E9%9D%A2%E8%BF%99%E4%B8%AA%E6%98%AF%E4%B8%AAm%E7%9A%842%2A2%E7%9F%A9%E9%98%B5.%281%29+For+general+h%2Cdiagonalize+m.%282%29+Find+the+projectors+to+the+eigenvalues+of+m+%283%29+What+happens+with+eigenvectors+when+h+%3D+What+happens+with+m+when+h+%3D)
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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