Linearly dependent set in vector space $P_4$

Question

Let $a \neq 0$. If the set $\{a + bx, ax +bx^2, b + ax^3\}$ is linearly dependent in vector space $P_4$ (polynomial with degree 4 in $\mathbb{R}$) , $a$ and $b$ must satisfy a relation....

By looking at my textbook, I try to form $P_4$ as a linear combination of the set elements,. $P_4 = k_1(a + bx) + k_2(ax+bx^2)+k_3(b+ax^3)$

Then, I need to form the correspondent coefficient of the polynomial, so there will be a linear system later, but I have a trouble in this state. Do you have any idea?

Answer 1

The given polynomials belong to the space of polynomials of degree at most $3$, which has $\{1,x,x^2,x^3\}$ as basis. The matrix having as columns the coordinates of the given polynomials with respect to this basis is $$ \begin{bmatrix} 1 & 0 & c \\ c & 1 & 0 \\ 0 & c & 0 \\ 0 & 0 & 1 \end{bmatrix} $$ where, since $a\ne0$, I considered the vectors multiplied by $a^{-1}$ and set $c=b/a$, which doesn't change linear dependence or linear independence.

This matrix has rank $3$, as Gaussian elimination goes as follows $$ \xrightarrow{R_2\gets R_2-cR_1} \begin{bmatrix} 1 & 0 & c \\ 0 & 1 & -c^2 \\ 0 & c & 0 \\ 0 & 0 & 1 \end{bmatrix} \xrightarrow{R_3\gets R_3-cR_2} \begin{bmatrix} 1 & 0 & c \\ 0 & 1 & -c^2 \\ 0 & 0 & c^3 \\ 0 & 0 & 1 \end{bmatrix} \xrightarrow{R_3\leftrightarrow R_4} \begin{bmatrix} 1 & 0 & c \\ 0 & 1 & -c^2 \\ 0 & 0 & 1 \\ 0 & 0 & c^3 \end{bmatrix} \xrightarrow{R_4\gets R_4-c^3R_3} \begin{bmatrix} 1 & 0 & c \\ 0 & 1 & -c^2 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} $$ Thus the given polynomials are linearly independent for every $a\ne0$ and every $b$.

If we allow $a=0$, the polynomials are, in this case, $bx$, $bx^2$ and $b$, which are again linearly independent unless $b=0$.

Thus the only case the polynomials are linearly dependent is the trivial one, that is, $a=b=0$.

Answer 2

I think there is no real number $a$ and $b$ with $a\neq0$, such that the set is linearly dependent. You need to remember the definition of a linearly dependent set. A set S is said to be linearly dependent, if there is a vector in S that can be expressed as a linear combination of the other vectors. Let us see the vectors in your question.

Let $v_1=a+bx$, $v_2=ax+bx^2$, and $v_3=b+ax^3$. Since $a\neq0$, then $v_3$ must be in degree 3. So we can't express $v_3$ as a linear combination of $v_1$ and $v_2$.

Now, there are two vectors remains, $v_1$ and $v_2$. We only need to check if a vector is multiple of the other vector or not. If $b=0$ then $v_1$ is a constant and $v_2$ in in degree 1, so no vector can be a multiple of the others. Similar thing happens when $b\neq0$.

So, whatever the value of $b$, the set is always linearly independent.