Why Does the Matrix Calculation Not Match Expected Results in Linear Mapping?

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SUMMARY

The discussion centers on the linear map T defined by T(p(x)) = p(−1) + ∫₀¹ p(x) dx, where R[x]_3 is the vector space of polynomials of degree at most 3. The matrix representation of T with respect to the standard basis is A = (2, -1/2, 4/3, -3/4). The kernel of T is spanned by the polynomials 1 + 4x, 2 - 3x², and 3 + 8x³. The confusion arises when multiplying the matrix A by the standard vector representation of p(x) = x, which correctly yields -1/2, confirming the linearity of T.

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Mathick
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Hello!

I don't know exactly how to state my question so I'll show you what my problem is.

Ex. Let T : [math]R[x]_3 →R[/math] be the function defined by [math] T(p(x)) = p(−1) + \int_{0}^{1} p(x) \,dx [/math], where [math]R[x]_3[/math] is a vector space of polynomials with degree at most 3. Show that $T$ is a linear map; write down the matrix of $T$ with respect to the standard bases of [math]R[x]_3[/math] and [math]R[/math] (the latter basis is of course given by [math] \left\{(1)\right\}[/math], i.e. the set consisting of the column vector of length 1 with entry 1), and find a basis for the kernel of $T$.

I found this matrix. It's a row vector [math]A=(2 \quad -\frac{1}{2} \quad\quad \frac{4}{3} \quad -\frac{3}{4})[/math]. And then the kernel basis is [math]1+4x,2-3x^2,3+8x^3[/math].

But I don't understand it. I mean if I multiply the matrix [math]A[/math] by a standard vector [math](1 \quad x \quad x^2 \quad x^3) [/math] I won't get, for example, for [math] p(x)=x [/math] the value [math] -\frac{1}{2}[/math].

Please, help me find the point where I misunderstand something. I really want to know what I am doing. I don't want to do maths like a robot without understanding.
 
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Mathick said:
Hello!

I don't know exactly how to state my question so I'll show you what my problem is.

Ex. Let T : [math]R[x]_3 →R[/math] be the function defined by [math] T(p(x)) = p(−1) + \int_{0}^{1} p(x) \,dx [/math], where [math]R[x]_3[/math] is a vector space of polynomials with degree at most 3. Show that $T$ is a linear map; write down the matrix of $T$ with respect to the standard bases of [math]R[x]_3[/math] and [math]R[/math] (the latter basis is of course given by [math] \left\{(1)\right\}[/math], i.e. the set consisting of the column vector of length 1 with entry 1), and find a basis for the kernel of $T$.

I found this matrix. It's a row vector [math]A=(2 \quad -\frac{1}{2} \quad\quad \frac{4}{3} \quad -\frac{3}{4})[/math]. And then the kernel basis is [math]1+4x,2-3x^2,3+8x^3[/math].

But I don't understand it. I mean if I multiply the matrix [math]A[/math] by a standard vector [math](1 \quad x \quad x^2 \quad x^3) [/math] I won't get, for example, for [math] p(x)=x [/math] the value [math] -\frac{1}{2}[/math].

Please, help me find the point where I misunderstand something. I really want to know what I am doing. I don't want to do maths like a robot without understanding.
With respect to the standard basis $\{1,x,x^2,x^3\}$ of $R[x]_3$, the polynomial $p(x) = a + bx + cx^2 + dx^3$ is represented by the vector $V = \begin{bmatrix}a\\b\\c\\d\end{bmatrix},$ so that $$T(p(x)) = AV = \begin{bmatrix}2&-\tfrac12 & \tfrac43 & -\tfrac34\end{bmatrix} \begin{bmatrix}a\\b\\c\\d\end{bmatrix}.$$ In particular, if $p(x) = x$ then $V = \begin{bmatrix}0\\1\\0\\0\end{bmatrix}$ and $T(p(x)) = \begin{bmatrix}2&-\tfrac12 & \tfrac43 & -\tfrac34\end{bmatrix} \begin{bmatrix}0\\1\\0\\0\end{bmatrix} = -\frac12.$
 

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