# Difficulty in linear algebra question.

Let $a_{ij}=a_ia_j$ $1\leq i,j\leq n$ where $a_1,... a_n$ are real numbers. Let $A=(a_{ij})$ be the $n \times n$ matrix . Then

1. It is spoosible to choose $a_1,... a_n$ so as to mke A non singular.

2. The matrix is positive definite if $(a_1,...,a_n)$ is a nonzero vector.

3. The matrix A is positive definite for all $(a_1,...,a_n)$.

4. For all $(a_1,...,a_n)$ zero is an eigen value of A.

My attempt:

Option 1 is false because if we choose $a_1=\frac{1}{\sqrt{2}} , a_2=\sqrt{2}$ then $2\times 2$ matrix $A$ is singular.

Option 2 is also false because if we take $a_1=1 \ a_2=2$ then $2\times2$ matrix $A$ is not positive definite. Option 3 is also false. But I am not getting option 4. Could you please help me? Thanks in advance.

• In all cases, the answer depends on whether $n>1$ (which is not given) or not. – Marc van Leeuwen May 20 '17 at 6:44
• You can also think of it as follows. All the rows of $A$ (actually also columns) are scalar multiples of $(a_1,a_2,\ldots,a_n)$. Therefore the rank of $A$ is _____ – Jyrki Lahtonen May 20 '17 at 6:45
• Your mode of reasoning for question 1 is erroneous. You confuse a "for all" for an "there exist". – Jean Marie May 20 '17 at 7:26
• A keyword for this kind of matrices: (symmetric) "rank one" matrices – Jean Marie May 20 '17 at 7:27

With these sorts of questions, it's good to write the matrix in a way that helps us think geometrically. So let us define $$\mathbf M = \begin{bmatrix} a_1 a_1 & \dots & a_1 a_n \\ \vdots & \ddots & \vdots \\ a_n a_1 & \dots & a_n a_n \end{bmatrix}, \ \ \ \ \ \mathbf a=\begin{bmatrix} a_1 \\ \vdots \\ a_n \end{bmatrix}.$$ I hope you can show that $$\mathbf M = \mathbf a \mathbf a^{\rm T},$$ where $^T$ denote taking a transpose.

Therefore, if $\mathbf x \in \mathbb R^n$ is any vector, we have $$\mathbf M \mathbf x = \mathbf a \mathbf a^{\rm T} \mathbf x = (\mathbf x . \mathbf a) \mathbf a,$$ where $.$ denotes taking the dot product. I encourage you to check this too.

The advantage of writing the matrix in this way is that taking dots products has some geometric meaning, and many of us have better intuition about shapes than about algebra.

Using your geometric understanding of dot products, it should be possible for you to verify that:

• $|\mathbf a |^2$ is an eigenvalue of $\mathbf M$. The corresponding eigenspace is the one-dimensional space of vectors $\mathbf x$ that are parallel to $\mathbf a$.

• $0$ is an eigenvalue of $\mathbf M$. The corresponding eigenspace is the $n-1$-dimensional space of vectors $\mathbf x$ that are perpendicular to $\mathbf a$.

I particularly encourage you to think about the case where $\mathbf a$ is a unit vector: here, you should be able to convince yourself that $\mathbf M = \mathbf a \mathbf a^{\rm T}$ represents the projection along the axis of $\mathbf a$. You can think about what the eigenvalues and eigenvectors mean geometrically in this case.

• P.S. For option 1, your conclusion is correct but your reasoning is not. You need to prove that any choice of $\mathbf a$ makes $\mathbf M = \mathbf a \mathbf a^{\rm T}$ singular. You only showed the matrix is singular for one choice of $\mathbf a$. Perhaps you can think about why having a zero eigenvalue means that the matrix is necessarily singular. – Kenny Wong May 20 '17 at 6:45
• that's a very good answer. – Hirakjyoti Das May 20 '17 at 7:05

As mentioned in the comments, I assume $n \ge 2$.

A key observation is that if $a$ denotes the vector $(a_1,\ldots,a_n)$, then $$A = aa^\top.$$

1) To prove 1. is true, you need to provide an example. However, to prove 1. is false, you need to show that all choices of $a_1,\ldots,a_n$ result in $A$ being singular; you cannot just provide one example. Hint: compute the rank of $A$. Alternatively, this follows from 4) if you prove 4) first.

2) and 3) To check positive definiteness you need to check $v^\top A v > 0$ for all vectors $v$. If you write $A$ as $aa^\top$, this condition becomes $(a^\top v)^2 > 0$. No matter what $a$ is, this condition will never hold for all $v$, since there is always some $v$ orthogonal to $a$. Your counterexample suffices to answer 2) and 3) as well. Alternatively, if you prove 4) first, then 2) and 3) follow easily since you will have a zero eigenvalue.

4) Hint: if $v$ is orthogonal to $a$, what is $Av$?