Compute $text{rk}(p)$ using Gauss reduction on $A$. Compute $dimbig(ker(p)big)$.
$begingroup$
Let $B = (1, X, X^2)$ be a basis for $mathbb{R}_2[X]$ and $p ∈ mathcal{L}big(mathbb{R}_2[X]big)$ be the linear map defined by $p(1) = frac{1}{3}(2 − X − X^2)$, $p(X) = frac{1}{3}(−1 + 2X − X^2)$ and $p(X^2) = frac{1}{3} (−1 − X + 2X^2)$.
By definition of the task, I found that $A=M_B(p)$, which is the matrix of
$p$ with respect to the basis $B$.
begin{bmatrix}
frac{2}{3} & -frac{1}{3} & -frac{1}{3} \
-frac{1}{3} & frac{2}{3} & -frac{1}{3} \
-frac{1}{3} & -frac{1}{3} &frac{2}{3} \
end{bmatrix}
Compute $text{rk}(p)$ using Gauss reduction on $A$. Compute $dimbig(ker(p)big)$.
I found $text{rk}(p)$ which is $2$. I looked at the definition of $ker$, and $dim(ker)$ but apparently I'm quite struggling to find it. How should I go about it?
linear-algebra matrices linear-transformations matrix-rank change-of-basis
edited Dec 15 '18 at 12:20
Batominovski
33k33293
asked Dec 13 '18 at 17:02
nasdaqnasdaq
62
$endgroup$
add a comment |
$begingroup$
Let $B = (1, X, X^2)$ be a basis for $mathbb{R}_2[X]$ and $p ∈ mathcal{L}big(mathbb{R}_2[X]big)$ be the linear map defined by $p(1) = frac{1}{3}(2 − X − X^2)$, $p(X) = frac{1}{3}(−1 + 2X − X^2)$ and $p(X^2) = frac{1}{3} (−1 − X + 2X^2)$.
By definition of the task, I found that $A=M_B(p)$, which is the matrix of
$p$ with respect to the basis $B$.
begin{bmatrix}
frac{2}{3} & -frac{1}{3} & -frac{1}{3} \
-frac{1}{3} & frac{2}{3} & -frac{1}{3} \
-frac{1}{3} & -frac{1}{3} &frac{2}{3} \
end{bmatrix}
Compute $text{rk}(p)$ using Gauss reduction on $A$. Compute $dimbig(ker(p)big)$.
I found $text{rk}(p)$ which is $2$. I looked at the definition of $ker$, and $dim(ker)$ but apparently I'm quite struggling to find it. How should I go about it?
linear-algebra matrices linear-transformations matrix-rank change-of-basis
edited Dec 15 '18 at 12:20
Batominovski
33k33293
asked Dec 13 '18 at 17:02
nasdaqnasdaq
62
$endgroup$
add a comment |
$begingroup$
Let $B = (1, X, X^2)$ be a basis for $mathbb{R}_2[X]$ and $p ∈ mathcal{L}big(mathbb{R}_2[X]big)$ be the linear map defined by $p(1) = frac{1}{3}(2 − X − X^2)$, $p(X) = frac{1}{3}(−1 + 2X − X^2)$ and $p(X^2) = frac{1}{3} (−1 − X + 2X^2)$.
By definition of the task, I found that $A=M_B(p)$, which is the matrix of
$p$ with respect to the basis $B$.
begin{bmatrix}
frac{2}{3} & -frac{1}{3} & -frac{1}{3} \
-frac{1}{3} & frac{2}{3} & -frac{1}{3} \
-frac{1}{3} & -frac{1}{3} &frac{2}{3} \
end{bmatrix}
Compute $text{rk}(p)$ using Gauss reduction on $A$. Compute $dimbig(ker(p)big)$.
I found $text{rk}(p)$ which is $2$. I looked at the definition of $ker$, and $dim(ker)$ but apparently I'm quite struggling to find it. How should I go about it?
linear-algebra matrices linear-transformations matrix-rank change-of-basis
edited Dec 15 '18 at 12:20
Batominovski
33k33293
asked Dec 13 '18 at 17:02
nasdaqnasdaq
62
$endgroup$
Let $B = (1, X, X^2)$ be a basis for $mathbb{R}_2[X]$ and $p ∈ mathcal{L}big(mathbb{R}_2[X]big)$ be the linear map defined by $p(1) = frac{1}{3}(2 − X − X^2)$, $p(X) = frac{1}{3}(−1 + 2X − X^2)$ and $p(X^2) = frac{1}{3} (−1 − X + 2X^2)$.
By definition of the task, I found that $A=M_B(p)$, which is the matrix of
$p$ with respect to the basis $B$.
begin{bmatrix}
frac{2}{3} & -frac{1}{3} & -frac{1}{3} \
-frac{1}{3} & frac{2}{3} & -frac{1}{3} \
-frac{1}{3} & -frac{1}{3} &frac{2}{3} \
end{bmatrix}
Compute $text{rk}(p)$ using Gauss reduction on $A$. Compute $dimbig(ker(p)big)$.
I found $text{rk}(p)$ which is $2$. I looked at the definition of $ker$, and $dim(ker)$ but apparently I'm quite struggling to find it. How should I go about it?
linear-algebra matrices linear-transformations matrix-rank change-of-basis
linear-algebra matrices linear-transformations matrix-rank change-of-basis
edited Dec 15 '18 at 12:20
Batominovski
33k33293
asked Dec 13 '18 at 17:02
nasdaqnasdaq
62
edited Dec 15 '18 at 12:20
Batominovski
33k33293
asked Dec 13 '18 at 17:02
nasdaqnasdaq
62
edited Dec 15 '18 at 12:20
Batominovski
33k33293
edited Dec 15 '18 at 12:20
Batominovski
33k33293
edited Dec 15 '18 at 12:20
Batominovski
33k33293
33k33293
asked Dec 13 '18 at 17:02
nasdaqnasdaq
62
asked Dec 13 '18 at 17:02
nasdaqnasdaq
62
asked Dec 13 '18 at 17:02
nasdaqnasdaq
62
62
add a comment |
add a comment |
1 Answer
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$begingroup$
There is a theorem called the rank-nullity theorem, which states that for a linear transformation $p:Vto W, text{rk}(p)+dim(ker(p))=dim(V)$. In your case, $dim(ker(p))=3-text{rk}(p)=3-2=1$.
answered Dec 13 '18 at 21:30
Shubham JohriShubham Johri
5,172717
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
There is a theorem called the rank-nullity theorem, which states that for a linear transformation $p:Vto W, text{rk}(p)+dim(ker(p))=dim(V)$. In your case, $dim(ker(p))=3-text{rk}(p)=3-2=1$.
answered Dec 13 '18 at 21:30
Shubham JohriShubham Johri
5,172717
$endgroup$
add a comment |
$begingroup$
There is a theorem called the rank-nullity theorem, which states that for a linear transformation $p:Vto W, text{rk}(p)+dim(ker(p))=dim(V)$. In your case, $dim(ker(p))=3-text{rk}(p)=3-2=1$.
answered Dec 13 '18 at 21:30
Shubham JohriShubham Johri
5,172717
$endgroup$
add a comment |
$begingroup$
There is a theorem called the rank-nullity theorem, which states that for a linear transformation $p:Vto W, text{rk}(p)+dim(ker(p))=dim(V)$. In your case, $dim(ker(p))=3-text{rk}(p)=3-2=1$.
answered Dec 13 '18 at 21:30
Shubham JohriShubham Johri
5,172717
$endgroup$
There is a theorem called the rank-nullity theorem, which states that for a linear transformation $p:Vto W, text{rk}(p)+dim(ker(p))=dim(V)$. In your case, $dim(ker(p))=3-text{rk}(p)=3-2=1$.
answered Dec 13 '18 at 21:30
Shubham JohriShubham Johri
5,172717
answered Dec 13 '18 at 21:30
Shubham JohriShubham Johri
5,172717
answered Dec 13 '18 at 21:30
Shubham JohriShubham Johri
5,172717
answered Dec 13 '18 at 21:30
Shubham JohriShubham Johri
5,172717
5,172717
add a comment |
add a comment |
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