Finding matrix given eigenvalues and eigenvectors.












2












$begingroup$


a) Let $B$ be a 2x2 symmetric matrix and let $u$, $v$ be two eigenvectors of $B$ associated with the eigenvalues $w$ and $l$ respectively. Suppose $u$ = $$ left[
begin{array}{c}
1/sqrt2\
-1/sqrt2
end{array}
right] $$
and $v$ = $$ left[
begin{array}{c}
1/sqrt2\
1/sqrt2
end{array}
right] $$
with $w$= 1 and $l$=3 respectively, find matrix $B$.



b) Let $C$ be another symmetric matrix of order n with a characteristic polynomial $(w-w_1)(w-w_2)...(w-w_n)$ where $w_1≤w_2≤...≤w_n$. Prove that for any nonzero vector $x$ that belongs in $R^n$, $w_1≤x^TCx/x^Tx≤w_n$.



What I have done:
For part (a), converting $u$ and $v$ to orthogonal basis, I get $u_o$=$$ left[
begin{array}{c}
1\
-1
end{array}
right] $$ and $v_o$=$$ left[
begin{array}{c}
1\
1
end{array}
right] $$. Letting $M$ = $$ left[
begin{array}{cc}
1&0\
0&3
end{array}
right] $$ and $S$ = $$ left[
begin{array}{cc}
1&1\
-1&1
end{array}
right] $$, the matrix B is obtained by putting together $SMS^{-1}$ = $$ left[
begin{array}{cc}
2&1\
1&2
end{array}
right] $$.



However, for part (b), while I can derive that $w_1,w_2,...,w_n$ are eigenvalues of C, I am unsure of how to proceed, hence may I get some help for this?






linear-algebra matrices eigenvalues-eigenvectors linear-transformations







share|cite|improve this question









$endgroup$












  • $begingroup$
    Use $Av=lambda v$ to get system of linear equations and solve.
    $endgroup$
    – Yadati Kiran
    Nov 26 '18 at 17:13
















2












$begingroup$


a) Let $B$ be a 2x2 symmetric matrix and let $u$, $v$ be two eigenvectors of $B$ associated with the eigenvalues $w$ and $l$ respectively. Suppose $u$ = $$ left[
begin{array}{c}
1/sqrt2\
-1/sqrt2
end{array}
right] $$
and $v$ = $$ left[
begin{array}{c}
1/sqrt2\
1/sqrt2
end{array}
right] $$
with $w$= 1 and $l$=3 respectively, find matrix $B$.



b) Let $C$ be another symmetric matrix of order n with a characteristic polynomial $(w-w_1)(w-w_2)...(w-w_n)$ where $w_1≤w_2≤...≤w_n$. Prove that for any nonzero vector $x$ that belongs in $R^n$, $w_1≤x^TCx/x^Tx≤w_n$.



What I have done:
For part (a), converting $u$ and $v$ to orthogonal basis, I get $u_o$=$$ left[
begin{array}{c}
1\
-1
end{array}
right] $$ and $v_o$=$$ left[
begin{array}{c}
1\
1
end{array}
right] $$. Letting $M$ = $$ left[
begin{array}{cc}
1&0\
0&3
end{array}
right] $$ and $S$ = $$ left[
begin{array}{cc}
1&1\
-1&1
end{array}
right] $$, the matrix B is obtained by putting together $SMS^{-1}$ = $$ left[
begin{array}{cc}
2&1\
1&2
end{array}
right] $$.



However, for part (b), while I can derive that $w_1,w_2,...,w_n$ are eigenvalues of C, I am unsure of how to proceed, hence may I get some help for this?






linear-algebra matrices eigenvalues-eigenvectors linear-transformations







share|cite|improve this question









$endgroup$












  • $begingroup$
    Use $Av=lambda v$ to get system of linear equations and solve.
    $endgroup$
    – Yadati Kiran
    Nov 26 '18 at 17:13














2












2








2


1



$begingroup$


a) Let $B$ be a 2x2 symmetric matrix and let $u$, $v$ be two eigenvectors of $B$ associated with the eigenvalues $w$ and $l$ respectively. Suppose $u$ = $$ left[
begin{array}{c}
1/sqrt2\
-1/sqrt2
end{array}
right] $$
and $v$ = $$ left[
begin{array}{c}
1/sqrt2\
1/sqrt2
end{array}
right] $$
with $w$= 1 and $l$=3 respectively, find matrix $B$.



b) Let $C$ be another symmetric matrix of order n with a characteristic polynomial $(w-w_1)(w-w_2)...(w-w_n)$ where $w_1≤w_2≤...≤w_n$. Prove that for any nonzero vector $x$ that belongs in $R^n$, $w_1≤x^TCx/x^Tx≤w_n$.



What I have done:
For part (a), converting $u$ and $v$ to orthogonal basis, I get $u_o$=$$ left[
begin{array}{c}
1\
-1
end{array}
right] $$ and $v_o$=$$ left[
begin{array}{c}
1\
1
end{array}
right] $$. Letting $M$ = $$ left[
begin{array}{cc}
1&0\
0&3
end{array}
right] $$ and $S$ = $$ left[
begin{array}{cc}
1&1\
-1&1
end{array}
right] $$, the matrix B is obtained by putting together $SMS^{-1}$ = $$ left[
begin{array}{cc}
2&1\
1&2
end{array}
right] $$.



However, for part (b), while I can derive that $w_1,w_2,...,w_n$ are eigenvalues of C, I am unsure of how to proceed, hence may I get some help for this?






linear-algebra matrices eigenvalues-eigenvectors linear-transformations







share|cite|improve this question









$endgroup$




a) Let $B$ be a 2x2 symmetric matrix and let $u$, $v$ be two eigenvectors of $B$ associated with the eigenvalues $w$ and $l$ respectively. Suppose $u$ = $$ left[
begin{array}{c}
1/sqrt2\
-1/sqrt2
end{array}
right] $$
and $v$ = $$ left[
begin{array}{c}
1/sqrt2\
1/sqrt2
end{array}
right] $$
with $w$= 1 and $l$=3 respectively, find matrix $B$.



b) Let $C$ be another symmetric matrix of order n with a characteristic polynomial $(w-w_1)(w-w_2)...(w-w_n)$ where $w_1≤w_2≤...≤w_n$. Prove that for any nonzero vector $x$ that belongs in $R^n$, $w_1≤x^TCx/x^Tx≤w_n$.



What I have done:
For part (a), converting $u$ and $v$ to orthogonal basis, I get $u_o$=$$ left[
begin{array}{c}
1\
-1
end{array}
right] $$ and $v_o$=$$ left[
begin{array}{c}
1\
1
end{array}
right] $$. Letting $M$ = $$ left[
begin{array}{cc}
1&0\
0&3
end{array}
right] $$ and $S$ = $$ left[
begin{array}{cc}
1&1\
-1&1
end{array}
right] $$, the matrix B is obtained by putting together $SMS^{-1}$ = $$ left[
begin{array}{cc}
2&1\
1&2
end{array}
right] $$.



However, for part (b), while I can derive that $w_1,w_2,...,w_n$ are eigenvalues of C, I am unsure of how to proceed, hence may I get some help for this?






linear-algebra matrices eigenvalues-eigenvectors linear-transformations




linear-algebra matrices eigenvalues-eigenvectors linear-transformations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 26 '18 at 17:06









Cheryl Cheryl

755




755












  • $begingroup$
    Use $Av=lambda v$ to get system of linear equations and solve.
    $endgroup$
    – Yadati Kiran
    Nov 26 '18 at 17:13


















  • $begingroup$
    Use $Av=lambda v$ to get system of linear equations and solve.
    $endgroup$
    – Yadati Kiran
    Nov 26 '18 at 17:13
















$begingroup$
Use $Av=lambda v$ to get system of linear equations and solve.
$endgroup$
– Yadati Kiran
Nov 26 '18 at 17:13




$begingroup$
Use $Av=lambda v$ to get system of linear equations and solve.
$endgroup$
– Yadati Kiran
Nov 26 '18 at 17:13










1 Answer
1






active

oldest

votes


















1












$begingroup$

Suppose that $lVert xrVert=1$. Then $x^Tx=1$. On the other hand, let $(v_1,ldots,v_n)$ an orthonormal basis of eigenvecors, such that the eigenvalu corresponding to $v_k$ is $w_k$. Then $x$ can be written as $alpha_1v_1+cdots+alpha_kv_k$. So, $Cx=alpha_1w_1v_1+cdots+alpha_nw_nv_n$ and so$$x^TCX=lvertalpha_1rvert^2w_1+cdots+lvertalpha_nrvert^2w_n.$$Since $lvertalpha_1rvert^2+cdots+lvertalpha_nrvert^2=1$, this number is between $w_1$ and $w_n$.



If $x$ is an arbitrary non-zero vector, you can apply the previus result to $frac x{lVert xrVert}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sir Why in particular is $lvertalpha_1rvert^2+cdots+lvertalpha_nrvert^2=1$?
    $endgroup$
    – Yadati Kiran
    Nov 26 '18 at 17:29










  • $begingroup$
    Because the basis is orthonormal and $lVert xrVert=1$.
    $endgroup$
    – José Carlos Santos
    Nov 26 '18 at 17:30












  • $begingroup$
    Oh ok sir. got it !
    $endgroup$
    – Yadati Kiran
    Nov 26 '18 at 17:31











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1 Answer
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1 Answer
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active

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1












$begingroup$

Suppose that $lVert xrVert=1$. Then $x^Tx=1$. On the other hand, let $(v_1,ldots,v_n)$ an orthonormal basis of eigenvecors, such that the eigenvalu corresponding to $v_k$ is $w_k$. Then $x$ can be written as $alpha_1v_1+cdots+alpha_kv_k$. So, $Cx=alpha_1w_1v_1+cdots+alpha_nw_nv_n$ and so$$x^TCX=lvertalpha_1rvert^2w_1+cdots+lvertalpha_nrvert^2w_n.$$Since $lvertalpha_1rvert^2+cdots+lvertalpha_nrvert^2=1$, this number is between $w_1$ and $w_n$.



If $x$ is an arbitrary non-zero vector, you can apply the previus result to $frac x{lVert xrVert}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sir Why in particular is $lvertalpha_1rvert^2+cdots+lvertalpha_nrvert^2=1$?
    $endgroup$
    – Yadati Kiran
    Nov 26 '18 at 17:29










  • $begingroup$
    Because the basis is orthonormal and $lVert xrVert=1$.
    $endgroup$
    – José Carlos Santos
    Nov 26 '18 at 17:30












  • $begingroup$
    Oh ok sir. got it !
    $endgroup$
    – Yadati Kiran
    Nov 26 '18 at 17:31
















1












$begingroup$

Suppose that $lVert xrVert=1$. Then $x^Tx=1$. On the other hand, let $(v_1,ldots,v_n)$ an orthonormal basis of eigenvecors, such that the eigenvalu corresponding to $v_k$ is $w_k$. Then $x$ can be written as $alpha_1v_1+cdots+alpha_kv_k$. So, $Cx=alpha_1w_1v_1+cdots+alpha_nw_nv_n$ and so$$x^TCX=lvertalpha_1rvert^2w_1+cdots+lvertalpha_nrvert^2w_n.$$Since $lvertalpha_1rvert^2+cdots+lvertalpha_nrvert^2=1$, this number is between $w_1$ and $w_n$.



If $x$ is an arbitrary non-zero vector, you can apply the previus result to $frac x{lVert xrVert}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sir Why in particular is $lvertalpha_1rvert^2+cdots+lvertalpha_nrvert^2=1$?
    $endgroup$
    – Yadati Kiran
    Nov 26 '18 at 17:29










  • $begingroup$
    Because the basis is orthonormal and $lVert xrVert=1$.
    $endgroup$
    – José Carlos Santos
    Nov 26 '18 at 17:30












  • $begingroup$
    Oh ok sir. got it !
    $endgroup$
    – Yadati Kiran
    Nov 26 '18 at 17:31














1












1








1





$begingroup$

Suppose that $lVert xrVert=1$. Then $x^Tx=1$. On the other hand, let $(v_1,ldots,v_n)$ an orthonormal basis of eigenvecors, such that the eigenvalu corresponding to $v_k$ is $w_k$. Then $x$ can be written as $alpha_1v_1+cdots+alpha_kv_k$. So, $Cx=alpha_1w_1v_1+cdots+alpha_nw_nv_n$ and so$$x^TCX=lvertalpha_1rvert^2w_1+cdots+lvertalpha_nrvert^2w_n.$$Since $lvertalpha_1rvert^2+cdots+lvertalpha_nrvert^2=1$, this number is between $w_1$ and $w_n$.



If $x$ is an arbitrary non-zero vector, you can apply the previus result to $frac x{lVert xrVert}$.






share|cite|improve this answer









$endgroup$



Suppose that $lVert xrVert=1$. Then $x^Tx=1$. On the other hand, let $(v_1,ldots,v_n)$ an orthonormal basis of eigenvecors, such that the eigenvalu corresponding to $v_k$ is $w_k$. Then $x$ can be written as $alpha_1v_1+cdots+alpha_kv_k$. So, $Cx=alpha_1w_1v_1+cdots+alpha_nw_nv_n$ and so$$x^TCX=lvertalpha_1rvert^2w_1+cdots+lvertalpha_nrvert^2w_n.$$Since $lvertalpha_1rvert^2+cdots+lvertalpha_nrvert^2=1$, this number is between $w_1$ and $w_n$.



If $x$ is an arbitrary non-zero vector, you can apply the previus result to $frac x{lVert xrVert}$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 26 '18 at 17:20









José Carlos SantosJosé Carlos Santos

156k22125227




156k22125227












  • $begingroup$
    Sir Why in particular is $lvertalpha_1rvert^2+cdots+lvertalpha_nrvert^2=1$?
    $endgroup$
    – Yadati Kiran
    Nov 26 '18 at 17:29










  • $begingroup$
    Because the basis is orthonormal and $lVert xrVert=1$.
    $endgroup$
    – José Carlos Santos
    Nov 26 '18 at 17:30












  • $begingroup$
    Oh ok sir. got it !
    $endgroup$
    – Yadati Kiran
    Nov 26 '18 at 17:31


















  • $begingroup$
    Sir Why in particular is $lvertalpha_1rvert^2+cdots+lvertalpha_nrvert^2=1$?
    $endgroup$
    – Yadati Kiran
    Nov 26 '18 at 17:29










  • $begingroup$
    Because the basis is orthonormal and $lVert xrVert=1$.
    $endgroup$
    – José Carlos Santos
    Nov 26 '18 at 17:30












  • $begingroup$
    Oh ok sir. got it !
    $endgroup$
    – Yadati Kiran
    Nov 26 '18 at 17:31
















$begingroup$
Sir Why in particular is $lvertalpha_1rvert^2+cdots+lvertalpha_nrvert^2=1$?
$endgroup$
– Yadati Kiran
Nov 26 '18 at 17:29




$begingroup$
Sir Why in particular is $lvertalpha_1rvert^2+cdots+lvertalpha_nrvert^2=1$?
$endgroup$
– Yadati Kiran
Nov 26 '18 at 17:29












$begingroup$
Because the basis is orthonormal and $lVert xrVert=1$.
$endgroup$
– José Carlos Santos
Nov 26 '18 at 17:30






$begingroup$
Because the basis is orthonormal and $lVert xrVert=1$.
$endgroup$
– José Carlos Santos
Nov 26 '18 at 17:30














$begingroup$
Oh ok sir. got it !
$endgroup$
– Yadati Kiran
Nov 26 '18 at 17:31




$begingroup$
Oh ok sir. got it !
$endgroup$
– Yadati Kiran
Nov 26 '18 at 17:31


















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