How can we find the orthonormal basis of $\{(x_1,\dots,x_n) \in \Bbb R^n \mid x_1+x_2+\cdots+x_n=0\}$?   It is easy to find a basis, but using Gram-Schmidt procedure seems difficult to obtain an orthonormal one.
123519
Find an orthonormal basis for the subspace of $\mathbb{R}^4$ that consists of vectors  perpendicular to $u = (1, -1, -1, 1)$.  I know the components of the vector $u$ is $u_1 = 1, u_2 = -1, u_3 = -1, u_4 = 1$.  I managed to find a vector $v$ that is perpendicular to the $u$.  This is done by $u_1v_1+ u_2v_2+ u_3v_3 + u_4v_4 = 0$.  This means $v_1 = -1, v_2 = 1, v_3= 1, v_4 = -1$  because $(1)(-1) + (-1)(1) + (-1)(1) + (1)(-1) = 0$.  So I found a vector $v$ perpendicular to the given vector $u$ which is $(-1, 1, 1, -1)$.  So my question would I be able to call $(-1, 1, 1, -1)$ my basis and perform Gram Schmidt process on it?
102251
How would you find the bases for   $W_1 = \{(a_1,a_2,a_3,a_4,a_5) \in F^5: a_1 - a_3 - a_4 = 0 \}$ and $W_2 = \{ (a_1,a_2,a_3,a_4,a_5) \in F^5: a_2 = a_3 = a_4$ and $a_1+a_5 = 0 \}$ ?  Would the basis for $W_1$ be $\{ (1 \,0 \, 1 \, 0 \, 0), (1 \, 0 \, 0 \, 1 \, 0) \}$ and $dim(W_1) = 2$?
105560
Let $U=M_{n}(\mathbb{C})$, and we define a bilinear form $\xi (A,B)=n\cdot tr(AB)-tr(A)tr(B)$. How do I find $dim(U_{\perp })$?  I know that $U_{\perp }=\{A \in M_{n}(\mathbb{C})|\ \forall B\in M_{n}(\mathbb{C}), n\cdot tr(AB)-tr(A)tr(B)=0 \}$. I have no idea how to continue. I'd appreciate any help. Thanks.
98622
given that $\{e_i\}_{n=1}^{\infty}$ is an orthonormal sequence in a hilbert space $H$, and $x\ne 0\in H$. Then could any one tell me which of the following is true?  $1.\lim_{n\to\infty}\langle x,e_n\rangle$ does not exist  $2.\lim_{n\to\infty}\langle x,e_n\rangle=\|x\|$  $3.\lim_{n\to\infty}\langle x,e_n\rangle=1$  $4.\lim_{n\to\infty}\langle x,e_n\rangle=0$  I just know one relations about orthonormal sequence in Hilbert space  $\sum_{n=1}^{\infty}|\langle x,e_i\rangle|^2\le \|x\|^2$. Thank you for help.
107718
If we have a subspace W of $\mathbb{R}^2$ spanned by $(3,4)$. Using the standard inner product, let E be the orthogonal projection of $\mathbb{R}^2$ onto W. Find an orthonormal basis in which E is represnted by the matrix: $\begin{bmatrix}     1      &amp; 0  \\     0      &amp; 0  \end{bmatrix}$  Now I figured out that $E(x_1,x_2) = \frac{1}{25}(9x_1+12x_2,12x_2+16x_2)$ and that the matrix for $E$ in the standard ordered basis is  $\frac{1}{25}\begin{bmatrix}     9      &amp; 12  \\    12      &amp; 16  \end{bmatrix}$. From here how would I figure out what the orthonormal basis is? I know that since the eigenvalues of $E$ are $\lambda =0,1$ then the basis for the other representation should be the unit eigenvectors. Would this be correct? It doesn't make sense to me because the orthonormal basis where the matrix representation of $E$ is the latter is just $\{(\frac{3}{5},\frac{4}{5})\}$ which does not span all of $\mathbb{R}^2$, but taking the eigenvectors would span all of $\mathbb{R}^2$.
67037
Why is it important that a basis be orthonormal? The requirement for Orthonormal basis is so often repeated in linear algebra that it seems linear algebra depends on it as requirement. What is gained (or lost) if the basis are not orthonormal? Is it just for convention and convenience or is it just to restrict to the type of problems that assumption of orthonormality is beneficial?
16411
Suppose that $A \in \mathbb{R}^{n\times n }$, and $x,y\in \mathbb{R}^n$.    Show that there exist numbers $a,b$ so that $\det{(A + sxy^*)}=a+bs$   Show that if $\det{(A)}\neq 0$ then $a = \det{(A)}$ and $b = \det{(A)} \cdot y^*A^{-1}x$.   Is it true that if $\det{(A)}=0$ then $a=0$?    My answer so far: The third question seems trivial and silly, and I don't understand why it's on there. I have solved the first one as follows: take $\beta$ to be any basis whose first element is $x$. Then with respect to basis $\beta$, $A+ sxy^*$ has the form   $$\left(  \begin{matrix} \alpha_{11} + sy_1 &amp; \alpha_{12} + sy_2  &amp; \dotsb &amp; \alpha_{1n}+sy_n \\ \alpha_{21}&amp;\dotsb&amp;&amp;\alpha_{2n} \\ \vdots &amp; \ddots &amp;&amp;\vdots\\ \alpha_{n1} &amp; \dotsb &amp;&amp;\alpha_{nn} \end{matrix} \right),$$  from which it's clear that the determinant is a polynomial in $s$ of degree $\leq 1$.    I am less clear on the second question.   For a limited case, note that if $A$ is diagonalizable and $x$ is an eigenvector of $A$ with eigenvalue $\lambda_x \neq 0$, then adding $sxy^*$ to $A$ has the effect of preserving all the eigenvectors, bumping up $\lambda_x$ to $\lambda_x + sy^*x$, and since the determinant is the product of all the eigenvalues, the product of $\lambda_1 \lambda_2  \dotsb (\lambda_x + sy^*x)\dotsb \lambda_n $ is   $$\det{(A)} + \det{(A)}\frac{sy^*x}{\lambda_x}=\det{(A)}\cdot sy^*(\frac{x}{\lambda_x}) = \det{(A)}\cdot s y^*(A^{-1}x).$$  Note that $y^*A^{-1}x$ is $\alpha_1 y_1 + \dotsb + \alpha_n y_n$, for $\alpha_i$ such that $x = \alpha_1 A_1 + \dotsb + \alpha_n A_n$, where $A_i$ are the columns of $A$.  But this is all I've got. Can someone give me a hint? I'm not looking for you to feed me the answer. Thanks as always
31677
If $B_1 = \{v_1,\ldots,v_n\}$ and $B_2 = \{v_1',\ldots,v_n'\}$ are basis for a vector space V, and $D_1= \{\delta v_1,\ldots, \delta v_n\}$ and $D_2 = \{\delta v_1',\ldots, \delta v_n'\}$ are the corresponding dual bases of $V^*$ prove that, if $P$ is the change-of-basis matrix from $B_1$ to $B_2$, then $(P^{-1})^T$ is the change-of-basis from $D_1$ to $D_2$  (this is a theorem from Schaum's outline of theory and problems of linear algebra)  My knowledge of linear algebra is not great, but so far I have that   \begin{align} \text{For any }v \in V, [v]_{B_1}&amp;=P[v]_{B_2}\\ \text{so } (\alpha_1,\ldots,\alpha_n)(v_1,\ldots,v_n)^T&amp;=P(\beta_1,\ldots,\beta_n)(v_1',\ldots,v_n')^T \end{align} for some $\alpha_1,\ldots,\alpha_n,\beta_1,\ldots,\beta_n \in F$ (where $F$ is a field)  How do I use the fact that $D_1$, $D_2$ are dual basis to further this proof? And also, where does the transpose of $P^{-1}$ come in? Can I use that $D_1 = \delta B_1$ and $D_2 = \delta B_2$ (where I am guessing $\delta$ is the kronecker delta?)  Any hints or suggestions are greatly appreciated, I know that my knowledge is quite limited with regard to this, links to text or sources that could explain a problem like this would be great.
60727
Let $A,B$ be real symmetric matrices, and for any $n\in \Bbb N^{+}$, and for all $x,y\in \Bbb R$, we have $$tr(xA+yB)^n=x^ntr(A^n)+y^ntr(B^n).$$  Show that $AB=0$.  My try:since $$(xA+yB)^n=x^nA^n+\binom{n}{1}x^{n-1}A^{n-1}yB+\cdots+\binom{n}{n-1}xA(yB)^{n-1}+y^nB^n$$ so $$tr\left(x^nA^n+\binom{n}{1}x^{n-1}A^{n-1}yB+\cdots+\binom{n}{n-1}xA(yB)^{n-1}+y^nB^n\right)=x^ntr(A^n)+y^ntr(B^n)$$  then  $$tr\left(\binom{n}{1}x^{n-1}A^{n-1}yB+\cdots+\binom{n}{n-1}xA(yB)^{n-1}\right)=0$$ so $$\binom{n}{1}x^{n-1}ytr(A^{n-1}B)+\cdots+\binom{n}{n-1}xy^{n-1}tr(AB^{n-1})=0$$ then I can't  I found this is a nice result, maybe this is an odd problem, and I can't solve it. Thank you.
101631
In the beginning of linear algebra courses, there are vectors in $\mathbb R^n$ and the dot product is introduced. We learn that if the dot product of two vectors is zero, then these vectors are called orthogonal and there is a right angle between them. Therefore, we understand perpendicularity from the definition of orthogonality.   In the last chapters, all previous notions are generalized as vector spaces and inner products are introduced. Then we learn vectors $\vec u$ and $\vec v$ are orthogonal if $\langle \vec u, \vec v \rangle=0$ in inner product spaces. Our previous knowledge guides us to seek a right angle between them.   Let $P$ be the vector space of first degree polynomials. Let $p(x)=ax+b$ and $q(x)=cx+d$. Define $\langle p(x), q(x) \rangle$ as  $$\langle p(x), q(x) \rangle=ac+bd$$  Then if we consider $p(x)=x-2$ and $q(x)=4x+2$, we find that $\langle p(x), q(x) \rangle=0$. But if we draw them, we see that there is $30.96$ degrees between them. So orthogonality in inner product spaces does not necessarily mean perpendicularity. If not perpendicularity, what should we understand from orthogonality in inner product spaces?
95457
Let $(M,g)$ be a Riemannian manifold of dimension $n$ with Riemannian connection $\nabla,$ and let $p \in M.$ Show that there exists a neighborhood $U \subset M$ of $p$ and $n$ (smooth) vector fields $E_1,...,E_n \in \chi (U),$ orthonormal at each point of $U,$ s.t. at $p,$ $\nabla_{E_i}E_j(p)=0.$  What I've found so far;  Let $V$ be a normal neighborhood at $p.$ Let $(W,x(x_1,...,x_n))$ be a local coordinate system at $p$ s.t. $x(W) \subset V.$  Let $\{X_i:=\frac{\partial}{\partial x_i}\}$ be the standard basis of $T_pM.$ By Gram-Schmidt, there exists an orthonormal basis $\{E_i\}$ of $T_pM.$ Consider the orthonormal frame $E_i(t),$ by parallel transporting $\{E_i(0)=E_i\}$ along a  a geodesic $\gamma :I \to M$ starting at $\gamma(0)=p$ and ending in $\gamma(1)=q,$ where $q \in V.$ This gives us an orthonormal frame at each point of $V.$ To satisfy the second condition, let $E_i(t)=\sum_{l}a_{li}X_l$ and  $E_j(t)=\sum_{s}b_{sj}(t)X_s.$ Then   $$\nabla_{E_i}E_j=\sum_{k}(\sum_{l,s} a_{li}(t)b_{sj}(t)\Gamma^k_{ls}+E_i(b_{kj}))X_k,$$  where $\Gamma^k_{is}$ are Christoffel symbols.   To have $\nabla_{E_i}E_j(p)=0,$ we must have $\sum_{l,s}a_{li}(0)b_{sj}(0)\Gamma^k_{ls}+E_i(b_{kj})(p)=0$ for each $k.$      How can I conclude the argument?
111637
Suppose $(e_1,e_2,...,e_n)$ is an orthonormal basis of the inner product space $V$ and $v_1,v_2,...,v_n$ are vectors of $V$ such that $$||e_j-v_j||&lt; \frac{1}{\sqrt{n}}$$ for each $j \in \left\{1,2,...,n \right \}$. Prove that $(v_1,v_2,...,v_n)$ is a basis of $V$.  I am completely lost and just starting to learn about inner product spaces. Could someone provide a proof with the explanation of how you got there?
115553
I would appreciate a hint on the following problem:     Let $V$ be a finite dimensional vector space over $F$. There are two scalar products such that: $$ \forall \ w,v \in V \ \Big(\langle v,w\rangle_1=0 \implies \langle v,w\rangle_2=0\Big) $$ Show that $$ \exists \ c \in F \ \ \forall \ w,v \in V \ \Big(\langle v,w\rangle_1=c \langle v,w\rangle_2\Big) $$   I have tried to define an orthonormal basis with respect to $\langle \cdot,\cdot\rangle_1$ hoping that the transformation matrices would be different by a constant, yet it seems to lead nowhere.
33926
Let $H$ be the inner product space = $\{f: \mathbb{R} \to \mathbb{C} \mid f \text{ is continuous and has period }2 \pi\}$ where the inner product is: $$\langle f,g \rangle = \int_{0}^{2\pi}f(t)\overline{g(t)} dt $$  How do I prove that for $n \in \mathbb{Z}$ and $e_n(t)=\dfrac{1}{\sqrt{2\cdot\pi}} e^{i n  t}$, $(e_n)_{n \in N}$ is a basis of $H$ (that it is orthonormal is easy).
7464
Given a set of vectors $v_1 \cdots v_n$, the $n\times n$ Gram matrix $G$ is defined as $G_{i,j}=v_i \cdot v_j$  Due to symmetry in the dot product, $G$ is Hermitian.  I'm trying to remember why $|G|=0$ iff the set of vectors are not linearly independent.
72298
Let $V$ be a finite dimensional real vector space with inner product $\langle \, , \rangle$ and let $W$ be a subspace of $V$. The orthogonal complement of $W$ is defined as   $$ W^\perp= \left\{ v \in V \,:\, \langle v,w \rangle = 0 \text{ for all } w \in W \right\}. $$   Prove the following: $\dim W + \dim W^\perp= \dim V$.   I'm not sure how to find the relationship between number of basis vectors in $W$ and $W^\perp$.
63164
I don't think I ever understood the rationale behind this.  I get that the dot product $\mathbf{a} \cdot  \mathbf{b} =\lVert \mathbf{a}\rVert \cdot\lVert \mathbf{b}\rVert \cos\theta$ is derived from the cosine rule. (Do correct me if I'm wrong.)  However, I never really understood why $\mathbf{a} \times  \mathbf{b} =\lVert \mathbf{a}\rVert \cdot\lVert \mathbf{b} \rVert \sin\theta$. Is there a proof for this?  In addition, how does the vector cross product lead to orthogonal vectors?  I remember learning that they do and I know how to solve equations using the respective formulas, but I never got why is it so.