tensor rank of an element in a tensor product











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Let $V$ and $W$ be finite-dimensional vector spaces over $k$ with $text{dim}(V)=n$ and $text{dim}(W)=m$.



How can I see that every element $t in V otimes_k W$ has tensor rank at most $text{min}{m,n}$.






abstract-algebra commutative-algebra tensor-products tensor-rank







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    up vote
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    down vote

    favorite












    Let $V$ and $W$ be finite-dimensional vector spaces over $k$ with $text{dim}(V)=n$ and $text{dim}(W)=m$.



    How can I see that every element $t in V otimes_k W$ has tensor rank at most $text{min}{m,n}$.






    abstract-algebra commutative-algebra tensor-products tensor-rank







    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Let $V$ and $W$ be finite-dimensional vector spaces over $k$ with $text{dim}(V)=n$ and $text{dim}(W)=m$.



      How can I see that every element $t in V otimes_k W$ has tensor rank at most $text{min}{m,n}$.






      abstract-algebra commutative-algebra tensor-products tensor-rank







      share|cite|improve this question













      Let $V$ and $W$ be finite-dimensional vector spaces over $k$ with $text{dim}(V)=n$ and $text{dim}(W)=m$.



      How can I see that every element $t in V otimes_k W$ has tensor rank at most $text{min}{m,n}$.






      abstract-algebra commutative-algebra tensor-products tensor-rank




      abstract-algebra commutative-algebra tensor-products tensor-rank






      share|cite|improve this question













      share|cite|improve this question











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      asked Nov 22 at 14:38









      idriskameni

      608




      608






















          2 Answers
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          down vote



          accepted










          Suppose rank $t=r$ and write
          $$
          t = sum_{i=1}^r v_i otimes w_i
          $$
          where $w_i in W$ and $v_i in V$. I claim that ${v_i}_{i=1}^r$ and ${w_i}_{i=1}^r$ must be linearly independent sets.



          Assume toward I contradiction that ${v_i}$ is not a linearly independent set. WLOG we can write $v_r=sum_{i=1}^{r-1} lambda_i v_i$. Observe that
          $$
          v_r otimes w_r = left(sum_{i=1}^{r-1} lambda_i v_iright) otimes w_i=sum_{i=1}^{r-1} left( v_i otimes lambda_i w_r right)
          $$
          Then we have
          $$
          t = v_r otimes w_r +sum_{i=1}^{r-1} v_i otimes w_i= sum_{i=1}^{r-1} v_i otimes lambda_i w_r+sum_{i=1}^{r-1} v_i otimes w_i = sum_{i=1}^{r-1} v_i otimes (w_i +lambda_i w_r),
          $$
          contradicting the fact that $t$ has rank $r$. So we must have ${v_i}_{i=1}^r$ is a linearly independent set. A near identical argument works for ${w_i}_{i=1}^r$. It follows that $dim V geq r$ and $dim W geq r$.






          share|cite|improve this answer





















          • Should I fix the basis ${v_i}_{i=1}^n$, and ${w_i}_{i=1}^m$ before? I mean, the elements of the sets $v_i$ that you have described are elements of the basis, right?
            – idriskameni
            Dec 4 at 12:04






          • 1




            Fixing this as a basis before hand would be implicitly assuming that the sets are linearly independent. We can show the sets are linearly independent without having set bases for V and W as linear independence is not something that depends on a basis. Once we get that they are linearly independent sets, the claim $dim V geq r$ and $dim W geq r$ are following from the fact that any linearly independent set can be extended to a basis. Nothing up til that point needs a basis, and bases are only used in a small way at the end.
            – Eric
            Dec 4 at 13:01


















          up vote
          1
          down vote













          Fix a basis $v_1, ldots, v_n$ of $V$. Then any tensor $t in V otimes W$ can be written $t = v_1 otimes w_1 + cdots + v_n otimes w_n$ for some vectors $w_i$ depending on $t$. This shows the tensor rank is at most $n$, and the argument to show that it is at most $m$ is identical.






          share|cite|improve this answer





















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            2 Answers
            2






            active

            oldest

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            2 Answers
            2






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            active

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            up vote
            2
            down vote



            accepted










            Suppose rank $t=r$ and write
            $$
            t = sum_{i=1}^r v_i otimes w_i
            $$
            where $w_i in W$ and $v_i in V$. I claim that ${v_i}_{i=1}^r$ and ${w_i}_{i=1}^r$ must be linearly independent sets.



            Assume toward I contradiction that ${v_i}$ is not a linearly independent set. WLOG we can write $v_r=sum_{i=1}^{r-1} lambda_i v_i$. Observe that
            $$
            v_r otimes w_r = left(sum_{i=1}^{r-1} lambda_i v_iright) otimes w_i=sum_{i=1}^{r-1} left( v_i otimes lambda_i w_r right)
            $$
            Then we have
            $$
            t = v_r otimes w_r +sum_{i=1}^{r-1} v_i otimes w_i= sum_{i=1}^{r-1} v_i otimes lambda_i w_r+sum_{i=1}^{r-1} v_i otimes w_i = sum_{i=1}^{r-1} v_i otimes (w_i +lambda_i w_r),
            $$
            contradicting the fact that $t$ has rank $r$. So we must have ${v_i}_{i=1}^r$ is a linearly independent set. A near identical argument works for ${w_i}_{i=1}^r$. It follows that $dim V geq r$ and $dim W geq r$.






            share|cite|improve this answer





















            • Should I fix the basis ${v_i}_{i=1}^n$, and ${w_i}_{i=1}^m$ before? I mean, the elements of the sets $v_i$ that you have described are elements of the basis, right?
              – idriskameni
              Dec 4 at 12:04






            • 1




              Fixing this as a basis before hand would be implicitly assuming that the sets are linearly independent. We can show the sets are linearly independent without having set bases for V and W as linear independence is not something that depends on a basis. Once we get that they are linearly independent sets, the claim $dim V geq r$ and $dim W geq r$ are following from the fact that any linearly independent set can be extended to a basis. Nothing up til that point needs a basis, and bases are only used in a small way at the end.
              – Eric
              Dec 4 at 13:01















            up vote
            2
            down vote



            accepted










            Suppose rank $t=r$ and write
            $$
            t = sum_{i=1}^r v_i otimes w_i
            $$
            where $w_i in W$ and $v_i in V$. I claim that ${v_i}_{i=1}^r$ and ${w_i}_{i=1}^r$ must be linearly independent sets.



            Assume toward I contradiction that ${v_i}$ is not a linearly independent set. WLOG we can write $v_r=sum_{i=1}^{r-1} lambda_i v_i$. Observe that
            $$
            v_r otimes w_r = left(sum_{i=1}^{r-1} lambda_i v_iright) otimes w_i=sum_{i=1}^{r-1} left( v_i otimes lambda_i w_r right)
            $$
            Then we have
            $$
            t = v_r otimes w_r +sum_{i=1}^{r-1} v_i otimes w_i= sum_{i=1}^{r-1} v_i otimes lambda_i w_r+sum_{i=1}^{r-1} v_i otimes w_i = sum_{i=1}^{r-1} v_i otimes (w_i +lambda_i w_r),
            $$
            contradicting the fact that $t$ has rank $r$. So we must have ${v_i}_{i=1}^r$ is a linearly independent set. A near identical argument works for ${w_i}_{i=1}^r$. It follows that $dim V geq r$ and $dim W geq r$.






            share|cite|improve this answer





















            • Should I fix the basis ${v_i}_{i=1}^n$, and ${w_i}_{i=1}^m$ before? I mean, the elements of the sets $v_i$ that you have described are elements of the basis, right?
              – idriskameni
              Dec 4 at 12:04






            • 1




              Fixing this as a basis before hand would be implicitly assuming that the sets are linearly independent. We can show the sets are linearly independent without having set bases for V and W as linear independence is not something that depends on a basis. Once we get that they are linearly independent sets, the claim $dim V geq r$ and $dim W geq r$ are following from the fact that any linearly independent set can be extended to a basis. Nothing up til that point needs a basis, and bases are only used in a small way at the end.
              – Eric
              Dec 4 at 13:01













            up vote
            2
            down vote



            accepted







            up vote
            2
            down vote



            accepted






            Suppose rank $t=r$ and write
            $$
            t = sum_{i=1}^r v_i otimes w_i
            $$
            where $w_i in W$ and $v_i in V$. I claim that ${v_i}_{i=1}^r$ and ${w_i}_{i=1}^r$ must be linearly independent sets.



            Assume toward I contradiction that ${v_i}$ is not a linearly independent set. WLOG we can write $v_r=sum_{i=1}^{r-1} lambda_i v_i$. Observe that
            $$
            v_r otimes w_r = left(sum_{i=1}^{r-1} lambda_i v_iright) otimes w_i=sum_{i=1}^{r-1} left( v_i otimes lambda_i w_r right)
            $$
            Then we have
            $$
            t = v_r otimes w_r +sum_{i=1}^{r-1} v_i otimes w_i= sum_{i=1}^{r-1} v_i otimes lambda_i w_r+sum_{i=1}^{r-1} v_i otimes w_i = sum_{i=1}^{r-1} v_i otimes (w_i +lambda_i w_r),
            $$
            contradicting the fact that $t$ has rank $r$. So we must have ${v_i}_{i=1}^r$ is a linearly independent set. A near identical argument works for ${w_i}_{i=1}^r$. It follows that $dim V geq r$ and $dim W geq r$.






            share|cite|improve this answer












            Suppose rank $t=r$ and write
            $$
            t = sum_{i=1}^r v_i otimes w_i
            $$
            where $w_i in W$ and $v_i in V$. I claim that ${v_i}_{i=1}^r$ and ${w_i}_{i=1}^r$ must be linearly independent sets.



            Assume toward I contradiction that ${v_i}$ is not a linearly independent set. WLOG we can write $v_r=sum_{i=1}^{r-1} lambda_i v_i$. Observe that
            $$
            v_r otimes w_r = left(sum_{i=1}^{r-1} lambda_i v_iright) otimes w_i=sum_{i=1}^{r-1} left( v_i otimes lambda_i w_r right)
            $$
            Then we have
            $$
            t = v_r otimes w_r +sum_{i=1}^{r-1} v_i otimes w_i= sum_{i=1}^{r-1} v_i otimes lambda_i w_r+sum_{i=1}^{r-1} v_i otimes w_i = sum_{i=1}^{r-1} v_i otimes (w_i +lambda_i w_r),
            $$
            contradicting the fact that $t$ has rank $r$. So we must have ${v_i}_{i=1}^r$ is a linearly independent set. A near identical argument works for ${w_i}_{i=1}^r$. It follows that $dim V geq r$ and $dim W geq r$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 22 at 14:53









            Eric

            2088




            2088












            • Should I fix the basis ${v_i}_{i=1}^n$, and ${w_i}_{i=1}^m$ before? I mean, the elements of the sets $v_i$ that you have described are elements of the basis, right?
              – idriskameni
              Dec 4 at 12:04






            • 1




              Fixing this as a basis before hand would be implicitly assuming that the sets are linearly independent. We can show the sets are linearly independent without having set bases for V and W as linear independence is not something that depends on a basis. Once we get that they are linearly independent sets, the claim $dim V geq r$ and $dim W geq r$ are following from the fact that any linearly independent set can be extended to a basis. Nothing up til that point needs a basis, and bases are only used in a small way at the end.
              – Eric
              Dec 4 at 13:01


















            • Should I fix the basis ${v_i}_{i=1}^n$, and ${w_i}_{i=1}^m$ before? I mean, the elements of the sets $v_i$ that you have described are elements of the basis, right?
              – idriskameni
              Dec 4 at 12:04






            • 1




              Fixing this as a basis before hand would be implicitly assuming that the sets are linearly independent. We can show the sets are linearly independent without having set bases for V and W as linear independence is not something that depends on a basis. Once we get that they are linearly independent sets, the claim $dim V geq r$ and $dim W geq r$ are following from the fact that any linearly independent set can be extended to a basis. Nothing up til that point needs a basis, and bases are only used in a small way at the end.
              – Eric
              Dec 4 at 13:01
















            Should I fix the basis ${v_i}_{i=1}^n$, and ${w_i}_{i=1}^m$ before? I mean, the elements of the sets $v_i$ that you have described are elements of the basis, right?
            – idriskameni
            Dec 4 at 12:04




            Should I fix the basis ${v_i}_{i=1}^n$, and ${w_i}_{i=1}^m$ before? I mean, the elements of the sets $v_i$ that you have described are elements of the basis, right?
            – idriskameni
            Dec 4 at 12:04




            1




            1




            Fixing this as a basis before hand would be implicitly assuming that the sets are linearly independent. We can show the sets are linearly independent without having set bases for V and W as linear independence is not something that depends on a basis. Once we get that they are linearly independent sets, the claim $dim V geq r$ and $dim W geq r$ are following from the fact that any linearly independent set can be extended to a basis. Nothing up til that point needs a basis, and bases are only used in a small way at the end.
            – Eric
            Dec 4 at 13:01




            Fixing this as a basis before hand would be implicitly assuming that the sets are linearly independent. We can show the sets are linearly independent without having set bases for V and W as linear independence is not something that depends on a basis. Once we get that they are linearly independent sets, the claim $dim V geq r$ and $dim W geq r$ are following from the fact that any linearly independent set can be extended to a basis. Nothing up til that point needs a basis, and bases are only used in a small way at the end.
            – Eric
            Dec 4 at 13:01










            up vote
            1
            down vote













            Fix a basis $v_1, ldots, v_n$ of $V$. Then any tensor $t in V otimes W$ can be written $t = v_1 otimes w_1 + cdots + v_n otimes w_n$ for some vectors $w_i$ depending on $t$. This shows the tensor rank is at most $n$, and the argument to show that it is at most $m$ is identical.






            share|cite|improve this answer

























              up vote
              1
              down vote













              Fix a basis $v_1, ldots, v_n$ of $V$. Then any tensor $t in V otimes W$ can be written $t = v_1 otimes w_1 + cdots + v_n otimes w_n$ for some vectors $w_i$ depending on $t$. This shows the tensor rank is at most $n$, and the argument to show that it is at most $m$ is identical.






              share|cite|improve this answer























                up vote
                1
                down vote










                up vote
                1
                down vote









                Fix a basis $v_1, ldots, v_n$ of $V$. Then any tensor $t in V otimes W$ can be written $t = v_1 otimes w_1 + cdots + v_n otimes w_n$ for some vectors $w_i$ depending on $t$. This shows the tensor rank is at most $n$, and the argument to show that it is at most $m$ is identical.






                share|cite|improve this answer












                Fix a basis $v_1, ldots, v_n$ of $V$. Then any tensor $t in V otimes W$ can be written $t = v_1 otimes w_1 + cdots + v_n otimes w_n$ for some vectors $w_i$ depending on $t$. This shows the tensor rank is at most $n$, and the argument to show that it is at most $m$ is identical.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 22 at 15:32









                Joppy

                5,573420




                5,573420






























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