Show that for M symmetric positive definite have $sup_{||x||=1}langle x, Mx rangle$=$lambda_1=:$ the largest...
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I know that if I can reach the step
$$sup_{||x||=1}langle x, Mx rangle=langle x,lambda_1 xrangle$$ then clearly $lambda_1$ follows. But how can we bridge this gap?
linear-algebra eigenvalues-eigenvectors
edited 19 hours ago
user587192
98410
asked 20 hours ago
Dan
827
add a comment |
up vote
0
down vote
favorite
I know that if I can reach the step
$$sup_{||x||=1}langle x, Mx rangle=langle x,lambda_1 xrangle$$ then clearly $lambda_1$ follows. But how can we bridge this gap?
linear-algebra eigenvalues-eigenvectors
edited 19 hours ago
user587192
98410
asked 20 hours ago
Dan
827
1
The key word to search is "Rayleigh quotient". See also: people.math.gatech.edu/~ecroot/notes_linear.pdf
– user587192
20 hours ago
Or alternatively, consider $varphi(t) = f(x + ty)$, where $f(x) = left(Mxmiddle|xright)/(x|x)$. Then $varphi(t)$ attains maximum at $t=0$, thus the differential calculus would work.
– xbh
20 hours ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I know that if I can reach the step
$$sup_{||x||=1}langle x, Mx rangle=langle x,lambda_1 xrangle$$ then clearly $lambda_1$ follows. But how can we bridge this gap?
linear-algebra eigenvalues-eigenvectors
edited 19 hours ago
user587192
98410
asked 20 hours ago
Dan
827
I know that if I can reach the step
$$sup_{||x||=1}langle x, Mx rangle=langle x,lambda_1 xrangle$$ then clearly $lambda_1$ follows. But how can we bridge this gap?
linear-algebra eigenvalues-eigenvectors
linear-algebra eigenvalues-eigenvectors
edited 19 hours ago
user587192
98410
asked 20 hours ago
Dan
827
edited 19 hours ago
user587192
98410
asked 20 hours ago
Dan
827
edited 19 hours ago
user587192
98410
edited 19 hours ago
user587192
98410
edited 19 hours ago
user587192
98410
98410
asked 20 hours ago
Dan
827
asked 20 hours ago
Dan
827
asked 20 hours ago
Dan
827
827
1
The key word to search is "Rayleigh quotient". See also: people.math.gatech.edu/~ecroot/notes_linear.pdf
– user587192
20 hours ago
Or alternatively, consider $varphi(t) = f(x + ty)$, where $f(x) = left(Mxmiddle|xright)/(x|x)$. Then $varphi(t)$ attains maximum at $t=0$, thus the differential calculus would work.
– xbh
20 hours ago
add a comment |
1
The key word to search is "Rayleigh quotient". See also: people.math.gatech.edu/~ecroot/notes_linear.pdf
– user587192
20 hours ago
Or alternatively, consider $varphi(t) = f(x + ty)$, where $f(x) = left(Mxmiddle|xright)/(x|x)$. Then $varphi(t)$ attains maximum at $t=0$, thus the differential calculus would work.
– xbh
20 hours ago
1
1
The key word to search is "Rayleigh quotient". See also: people.math.gatech.edu/~ecroot/notes_linear.pdf
– user587192
20 hours ago
The key word to search is "Rayleigh quotient". See also: people.math.gatech.edu/~ecroot/notes_linear.pdf
– user587192
20 hours ago
Or alternatively, consider $varphi(t) = f(x + ty)$, where $f(x) = left(Mxmiddle|xright)/(x|x)$. Then $varphi(t)$ attains maximum at $t=0$, thus the differential calculus would work.
– xbh
20 hours ago
Or alternatively, consider $varphi(t) = f(x + ty)$, where $f(x) = left(Mxmiddle|xright)/(x|x)$. Then $varphi(t)$ attains maximum at $t=0$, thus the differential calculus would work.
– xbh
20 hours ago
add a comment |
1 Answer
1
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By the spectral theorem we can take an $textit{orthonormal basis}$ of $mathbb R^n$ consisting of eigenvectors of $M$, say $v_1,ldots,v_n$ with corresponding eigenvalues $lambda_1,ldots,lambda_n$. Then, for any $Vert vVert=1$ we can write $v=c_1v_1+cdots+c_nv_n$. Try expanding $langle Mv,vrangle$ using linearity of the inner product, the fact that $langle v_i,v_jrangle=delta_{i,j}$, and the fact that $Mv_i=lambda_iv_i$ to show that $langle Mv,vrangleleqlambda^*$ where $lambda^*$ is your max eigenvalue. Then can you exhibit a specific $v$ with norm $1$ for which equality holds (think eigenvectors)?
answered 20 hours ago
Dave
8,16311033
1
Thank you. The orthogonal basis was the key. $||v||=1 Rightarrow sum_{i}^{n} c_i^2=1$ then write $langle v, Mv rangle= sum_i^n lambda_i c_i^2 leq sum_i^n lambda_1 c_i^2 = lambda_1$
– Dan
19 hours ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
By the spectral theorem we can take an $textit{orthonormal basis}$ of $mathbb R^n$ consisting of eigenvectors of $M$, say $v_1,ldots,v_n$ with corresponding eigenvalues $lambda_1,ldots,lambda_n$. Then, for any $Vert vVert=1$ we can write $v=c_1v_1+cdots+c_nv_n$. Try expanding $langle Mv,vrangle$ using linearity of the inner product, the fact that $langle v_i,v_jrangle=delta_{i,j}$, and the fact that $Mv_i=lambda_iv_i$ to show that $langle Mv,vrangleleqlambda^*$ where $lambda^*$ is your max eigenvalue. Then can you exhibit a specific $v$ with norm $1$ for which equality holds (think eigenvectors)?
answered 20 hours ago
Dave
8,16311033
1
Thank you. The orthogonal basis was the key. $||v||=1 Rightarrow sum_{i}^{n} c_i^2=1$ then write $langle v, Mv rangle= sum_i^n lambda_i c_i^2 leq sum_i^n lambda_1 c_i^2 = lambda_1$
– Dan
19 hours ago
add a comment |
up vote
0
down vote
accepted
By the spectral theorem we can take an $textit{orthonormal basis}$ of $mathbb R^n$ consisting of eigenvectors of $M$, say $v_1,ldots,v_n$ with corresponding eigenvalues $lambda_1,ldots,lambda_n$. Then, for any $Vert vVert=1$ we can write $v=c_1v_1+cdots+c_nv_n$. Try expanding $langle Mv,vrangle$ using linearity of the inner product, the fact that $langle v_i,v_jrangle=delta_{i,j}$, and the fact that $Mv_i=lambda_iv_i$ to show that $langle Mv,vrangleleqlambda^*$ where $lambda^*$ is your max eigenvalue. Then can you exhibit a specific $v$ with norm $1$ for which equality holds (think eigenvectors)?
answered 20 hours ago
Dave
8,16311033
1
Thank you. The orthogonal basis was the key. $||v||=1 Rightarrow sum_{i}^{n} c_i^2=1$ then write $langle v, Mv rangle= sum_i^n lambda_i c_i^2 leq sum_i^n lambda_1 c_i^2 = lambda_1$
– Dan
19 hours ago
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
By the spectral theorem we can take an $textit{orthonormal basis}$ of $mathbb R^n$ consisting of eigenvectors of $M$, say $v_1,ldots,v_n$ with corresponding eigenvalues $lambda_1,ldots,lambda_n$. Then, for any $Vert vVert=1$ we can write $v=c_1v_1+cdots+c_nv_n$. Try expanding $langle Mv,vrangle$ using linearity of the inner product, the fact that $langle v_i,v_jrangle=delta_{i,j}$, and the fact that $Mv_i=lambda_iv_i$ to show that $langle Mv,vrangleleqlambda^*$ where $lambda^*$ is your max eigenvalue. Then can you exhibit a specific $v$ with norm $1$ for which equality holds (think eigenvectors)?
answered 20 hours ago
Dave
8,16311033
By the spectral theorem we can take an $textit{orthonormal basis}$ of $mathbb R^n$ consisting of eigenvectors of $M$, say $v_1,ldots,v_n$ with corresponding eigenvalues $lambda_1,ldots,lambda_n$. Then, for any $Vert vVert=1$ we can write $v=c_1v_1+cdots+c_nv_n$. Try expanding $langle Mv,vrangle$ using linearity of the inner product, the fact that $langle v_i,v_jrangle=delta_{i,j}$, and the fact that $Mv_i=lambda_iv_i$ to show that $langle Mv,vrangleleqlambda^*$ where $lambda^*$ is your max eigenvalue. Then can you exhibit a specific $v$ with norm $1$ for which equality holds (think eigenvectors)?
answered 20 hours ago
Dave
8,16311033
answered 20 hours ago
Dave
8,16311033
answered 20 hours ago
Dave
8,16311033
answered 20 hours ago
Dave
8,16311033
8,16311033
1
Thank you. The orthogonal basis was the key. $||v||=1 Rightarrow sum_{i}^{n} c_i^2=1$ then write $langle v, Mv rangle= sum_i^n lambda_i c_i^2 leq sum_i^n lambda_1 c_i^2 = lambda_1$
– Dan
19 hours ago
add a comment |
1
Thank you. The orthogonal basis was the key. $||v||=1 Rightarrow sum_{i}^{n} c_i^2=1$ then write $langle v, Mv rangle= sum_i^n lambda_i c_i^2 leq sum_i^n lambda_1 c_i^2 = lambda_1$
– Dan
19 hours ago
1
1
Thank you. The orthogonal basis was the key. $||v||=1 Rightarrow sum_{i}^{n} c_i^2=1$ then write $langle v, Mv rangle= sum_i^n lambda_i c_i^2 leq sum_i^n lambda_1 c_i^2 = lambda_1$
– Dan
19 hours ago
Thank you. The orthogonal basis was the key. $||v||=1 Rightarrow sum_{i}^{n} c_i^2=1$ then write $langle v, Mv rangle= sum_i^n lambda_i c_i^2 leq sum_i^n lambda_1 c_i^2 = lambda_1$
– Dan
19 hours ago
add a comment |
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1
The key word to search is "Rayleigh quotient". See also: people.math.gatech.edu/~ecroot/notes_linear.pdf
– user587192
20 hours ago
Or alternatively, consider $varphi(t) = f(x + ty)$, where $f(x) = left(Mxmiddle|xright)/(x|x)$. Then $varphi(t)$ attains maximum at $t=0$, thus the differential calculus would work.
– xbh
20 hours ago